Ch 5  ·  Q–
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Chapter 5 Exercise 5.7 Solutions

Continuity and Differentiability

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 5 Exercise 5.7

Class 12 Mathematics Exercise 5.7 NCERT Solutions Continuity and Differentiability Class 12 Mathematics Chapter 5 CBSE Board Exam JEE Main CUET Higher Order Derivatives Second Order Derivatives Second Derivative Repeated Differentiation Implicit Differentiation Exponential Functions Logarithmic Functions Inverse Trigonometric Functions
17 Questions
40–55 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks

Find the second order derivative of

\[ x^{2}+3x+2 \]
📘 Concept & Theory
Concept/Theory
›

The second order derivative of a function is obtained by differentiating its first derivative once again with respect to the independent variable.

If

\[ y=f(x), \]
then the first derivative is
\[ \frac{dy}{dx}=f'(x). \]
Differentiating the first derivative again gives the second order derivative:
\[ \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^{2}y}{dx^{2}} = f''(x). \]
Thus, to find the second order derivative, we differentiate the given function twice.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the given expression be \(y\).

  2. Differentiate \(y\) with respect to \(x\) to obtain \(\dfrac{dy}{dx}\).

  3. Differentiate \(\dfrac{dy}{dx}\) once again with respect to \(x\).

  4. Write the resulting expression as \(\dfrac{d^{2}y}{dx^{2}}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  8 steps
  1. Let
    \[y=x^{2}+3x+2\]
  2. Differentiate both sides with respect to \(x\):
    \[ \frac{dy}{dx} = \frac{d}{dx}\left(x^{2}+3x+2\right) \]
  3. Using the power rule
    \[\frac{d}{dx}(x^{n})=nx^{n-1}\]
  4. we differentiate each term separately:
    \[\frac{d}{dx}(x^{2})=2x\]
    \[\frac{d}{dx}(3x)=3\]
    \[\frac{d}{dx}(2)=0\]
  5. Therefore,
    \[\frac{dy}{dx}=2x+3+0\]
    \[\boxed{\frac{dy}{dx}=2x+3}\]
  6. Now differentiate the first derivative once again with respect to \(x\):
    \[\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(2x+3\right)\]
  7. Differentiate each term:
    \[\frac{d}{dx}(2x)=2\]
    \[\frac{d}{dx}(3)=0\]
  8. Hence,
    \[\frac{d^{2}y}{dx^{2}}=2+0\]
    \[\boxed{\frac{d^{2}y}{dx^{2}}=2}\]
🎯 Exam Significance
Exam Significance
›
This question tests the fundamental procedure of finding higher order derivatives. Such questions are important because they establish the differentiation skills required for more advanced applications of derivatives. In board examinations, students should clearly show the first derivative before differentiating it again to obtain the second derivative. Writing the intermediate step also demonstrates the correct application of the power rule and reduces the possibility of calculation errors.
Significance for Competitive Entrance Examinations

Higher order derivatives frequently appear as intermediate steps in problems involving maxima and minima, concavity, approximations, differential equations, and functions defined through more complicated expressions. Although this particular question is elementary, mastering the basic process allows students to handle second and higher derivatives efficiently when the function involves products, quotients, composite functions, implicit functions, or parametric equations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. The second order derivative is obtained by differentiating the first derivative once again.

  2. If \(y=f(x)\), then the second derivative is represented by \(\dfrac{d^{2}y}{dx^{2}}\) or \(f''(x)\).

  3. For \(y=x^{2}+3x+2\), the first derivative is \(2x+3\).

  4. The second derivative is obtained by differentiating \(2x+3\).

  5. The constant term vanishes during differentiation because the derivative of a constant is zero.

  6. The final result is \(\boxed{\dfrac{d^{2}y}{dx^{2}}=2}\).
↑ Top
1 / 17  ·  6%
Q2 →
Q2
NUMERIC3 marks

Find the second order derivative of

\[ x^{20} \]
📘 Concept & Theory
Concept/Theory
›

The second order derivative is obtained by differentiating a function twice with respect to the independent variable. If

\[ y=f(x), \]
then
\[ \frac{dy}{dx}=f'(x) \]
is the first derivative, and differentiating it once again gives
\[ \frac{d^2y}{dx^2}=f''(x). \]

Here, the given function is a power function. Therefore, the power rule of differentiation is used:

\[ \frac{d}{dx}(x^n)=nx^{n-1}. \]
On differentiating a second time, the exponent is reduced by one again and the new exponent is multiplied by the existing coefficient.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the given expression be \(y=x^{20}\).

  2. Differentiate \(y\) once using the power rule to obtain \(\dfrac{dy}{dx}\).

  3. Differentiate the first derivative again using the power rule.

  4. Simplify the numerical coefficient \(20\times19\).

  5. Write the resulting expression as the second order derivative.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Let
    \[y=x^{20}\]
  2. Differentiate both sides with respect to \(x\):
    \[\frac{dy}{dx}=\frac{d}{dx}\left(x^{20}\right)\]
  3. Using the power rule
    \[\frac{d}{dx}(x^n)=nx^{n-1}\]
  4. we get
    \[\frac{dy}{dx}=20x^{20-1}\]
    \[\boxed{\frac{dy}{dx}=20x^{19}}\]
  5. Now differentiate the first derivative once again with respect to \(x\):
    \[\frac{d^2y}{dx^2}=\frac{d}{dx}\left(20x^{19}\right)\]
  6. Since \(20\) is a constant, it remains as a coefficient:
    \[\frac{d^2y}{dx^2}=20\frac{d}{dx}\left(x^{19}\right)\]
  7. Applying the power rule again:
    \[\frac{d}{dx}\left(x^{19}\right)=19x^{19-1}\]
  8. Therefore,
    \[\frac{d^2y}{dx^2}=20(19x^{18})\]
    \[\frac{d^2y}{dx^2}=20\times19x^{18}\]
  9. Now simplify the numerical coefficient:
    \[20\times19=380\]
  10. Hence,
    \[\boxed{\frac{d^2y}{dx^2}=380x^{18}}\]
🎯 Exam Significance
Exam Significance
›

This question reinforces the direct application of the power rule for finding higher order derivatives. In board examinations, students should show both differentiation steps clearly: first obtain \(\dfrac{dy}{dx}\), and then differentiate it to obtain \(\dfrac{d^2y}{dx^2}\). Keeping the coefficient \(20\) outside during the second differentiation helps avoid mistakes.

Significance for Competitive Entrance Examination Aspirants

Higher order derivatives are frequently used as components of more advanced calculus problems. The ability to differentiate power functions rapidly is particularly useful in problems involving maxima and minima, Taylor expansions, approximations, differential equations, and functions requiring repeated differentiation. Recognising the pattern of repeated power differentiation also improves calculation speed in competitive examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The power rule is \(\dfrac{d}{dx}(x^n)=nx^{n-1}\).

  2. For a second derivative, the power rule is applied twice.

  3. For \(y=x^{20}\), the first derivative is \(20x^{19}\).

  4. During the second differentiation, \(20\) remains as a constant coefficient.

  5. The second differentiation produces the additional factor \(19\).

  6. The coefficient becomes \(20\times19=380\).

  7. Therefore, \(\boxed{\dfrac{d^2y}{dx^2}=380x^{18}}\).
← Q1
2 / 17  ·  12%
Q3 →
Q3
NUMERIC3 marks

Find the second order derivative of

\[ x\cos x \]
📘 Concept & Theory
Concept/Theory
›

The given function is a product of two functions, \(x\) and \(\cos x\). Therefore, the product rule of differentiation is required.

The product rule states that if

\[ y=u\cdot v, \]

then

\[\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\]

Since the question asks for the second order derivative, the product rule must be applied first to obtain \(\dfrac{dy}{dx}\), and then applied again while differentiating the resulting expression.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(y=x\cos x\).

  2. Identify the two factors as \(u=x\) and \(v=\cos x\).

  3. Apply the product rule to find \(\dfrac{dy}{dx}\).

  4. Differentiate the first derivative term by term to obtain \(\dfrac{d^2y}{dx^2}\).

  5. Apply the product rule again to the term \(x\sin x\).

  6. Combine like terms and simplify the final result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Let
    \[y=x\cos x\]
  2. Since \(y\) is the product of \(x\) and \(\cos x\), use the product rule:
    \[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\]
  3. Here,
    \[u=x,\quad v=\cos x\]
  4. Therefore,
    \[\frac{du}{dx}=1\]
    and
    \[\frac{dv}{dx}=-\sin x\]
  5. Hence,
    \[\frac{dy}{dx}=x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(x)\]
    \[\frac{dy}{dx}=x(-\sin x)+\cos x(1)\]
    \[\frac{dy}{dx}=-x\sin x+\cos x\]
  6. Therefore,
    \[\boxed{\frac{dy}{dx}=\cos x-x\sin x}\]
  7. Now differentiate the first derivative again with respect to \(x\):
    \[\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\cos x-x\sin x\right)\]
  8. Differentiate each term separately:
    \[\frac{d}{dx}(\cos x)=-\sin x\]
  9. For the second term, we need to differentiate \(x\sin x\). Since it is a product, apply the product rule:
    \[\frac{d}{dx}(x\sin x)=x\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}(x)\]
    \[=x\cos x+\sin x\]
  10. Therefore,
    \[\frac{d^2y}{dx^2}=-\sin x-\left(x\cos x+\sin x\right)\]
    \[=-\sin x-x\cos x-\sin x\]
  11. Combining the two \(-\sin x\) terms:
    \[\boxed{\frac{d^2y}{dx^2}=-2\sin x-x\cos x}\]
🎯 Exam Significance
Exam Significance
›

This problem is important because it combines higher order differentiation with the product rule. A common error is to differentiate \(x\sin x\) as if only one factor were variable. Students should recognise that both \(x\) and \(\sin x\) depend on \(x\), so the product rule must be applied again. Showing both differentiation stages clearly is useful for obtaining full stepwise marks.

Significance for Competitive Entrance Examination Aspirants

Problems involving repeated differentiation of products of algebraic and trigonometric functions are common building blocks for more advanced calculus. Efficient application of the product rule is useful in questions involving higher derivatives, maxima and minima, Taylor and Maclaurin expansions, differential equations, and function analysis. The key skill is recognising the product structure immediately and applying the appropriate rule without expanding unnecessarily.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. When two functions are multiplied, use the product rule.

  2. The product rule is \(\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\).

  3. The first derivative of \(x\cos x\) is \(\cos x-x\sin x\).

  4. While finding the second derivative, the term \(x\sin x\) again requires the product rule.

  5. \(\dfrac{d}{dx}(\sin x)=\cos x\) and \(\dfrac{d}{dx}(\cos x)=-\sin x\).

  6. The two \(-\sin x\) terms combine to give \(-2\sin x\).

  7. The final result is \(\boxed{\dfrac{d^2y}{dx^2}=-2\sin x-x\cos x}\).
← Q2
3 / 17  ·  18%
Q4 →
Q4
NUMERIC3 marks

Find the second order derivative of

\[ \log x \]
📘 Concept & Theory
Concept/Theory
›

The second order derivative is obtained by differentiating the function twice with respect to \(x\). For the logarithmic function, we use

\[ \frac{d}{dx}(\log x)=\frac{1}{x}, \qquad x>0. \]

After finding the first derivative, we differentiate it once again. Since

\[ \frac{1}{x}=x^{-1}, \]
the power rule can be used conveniently:
\[ \frac{d}{dx}(x^n)=nx^{n-1}. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(y=\log x\).

  2. Differentiate \(\log x\) to obtain the first derivative.

  3. Rewrite \(\dfrac{1}{x}\) as \(x^{-1}\).

  4. Apply the power rule to \(x^{-1}\).

  5. Simplify the result and express the second derivative in fractional form.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Let
    \[y=\log x\]
  2. Differentiate both sides with respect to \(x\):
    \[\frac{dy}{dx}=\frac{d}{dx}(\log x)\]
  3. Using the standard derivative
    \[\frac{d}{dx}(\log x)=\frac{1}{x},\quad x>0,\]
  4. we get
    \[\boxed{\frac{dy}{dx}=\frac{1}{x}}\]
  5. Rewrite the first derivative using a negative exponent:
    \[\frac{dy}{dx}=x^{-1}\]
  6. Now differentiate again with respect to \(x\):
    \[\frac{d^2y}{dx^2}=\frac{d}{dx}\left(x^{-1}\right)\]
  7. Using the power rule
    \[\frac{d}{dx}(x^n)=nx^{n-1}\]
  8. with \(n=-1\), we obtain
    \[\frac{d^2y}{dx^2}=(-1)x^{-1-1}\]
    \[=-x^{-2}\]
  9. Using
    \[x^{-2}=\frac{1}{x^2}\]
  10. we get
    \[\boxed{\frac{d^2y}{dx^2}=-\frac{1}{x^2}}\]
🎯 Exam Significance
Exam Significance
›

This question tests the direct differentiation of a logarithmic function followed by a second differentiation using the power rule. It is important to show the transition from \(\dfrac{1}{x}\) to \(x^{-1}\), because this makes the application of the power rule transparent. The restriction \(x>0\) should also be remembered for the real-valued function \(\log x\).

Significance for Competitive Entrance Examination Aspirants

The derivative of \(\log x\) and its higher derivatives occur frequently in calculus problems involving logarithmic functions, monotonicity, maxima and minima, approximation, Taylor and Maclaurin expansions, and differential equations. Recognising that \(\dfrac{1}{x}\) is a power function allows the second derivative to be obtained quickly and accurately.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. For the real-valued function \(\log x\), the domain is \(x>0\).

  2. The first derivative is \(\dfrac{d}{dx}(\log x)=\dfrac{1}{x}\).

  3. The expression \(\dfrac{1}{x}\) can be written as \(x^{-1}\).

  4. Applying the power rule to \(x^{-1}\) gives \(-x^{-2}\).

  5. The negative exponent can be converted into fractional form: \(x^{-2}=\dfrac{1}{x^2}\).

  6. Therefore, \(\boxed{\dfrac{d^2y}{dx^2}=-\dfrac{1}{x^2}}\).
← Q3
4 / 17  ·  24%
Q5 →
Q5
NUMERIC3 marks

Find the second order derivative of

\[ x^{3}\log x \]
📘 Concept & Theory
Concept/Theory
›

The given function is the product of two functions, \(x^3\) and \(\log x\). Therefore, the product rule of differentiation is required.

If

\[ y=u\cdot v, \]

then

\[\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\]

Since a second order derivative is required, the resulting first derivative must be differentiated once again. The product rule will be required again because the first derivative contains the product \(x^2\log x\).

We also use the standard derivatives

\[ \frac{d}{dx}(x^n)=nx^{n-1} \]
and
\[ \frac{d}{dx}(\log x)=\frac{1}{x}, \qquad x>0. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(y=x^3\log x\).

  2. Identify \(x^3\) and \(\log x\) as the two factors.

  3. Apply the product rule to find \(\dfrac{dy}{dx}\).

  4. Simplify the first derivative.

  5. Differentiate the first derivative term by term.

  6. Apply the product rule again to \(x^2\log x\).

  7. Combine like terms and factor the final result where appropriate.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. Let
    \[y=x^3\log x\]
  2. Since \(y\) is a product of \(x^3\) and \(\log x\), use the product rule:
    \[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\]
  3. Take
    \[u=x^3,\quad v=\log x\]
  4. Then
    \[\frac{du}{dx}=3x^2\]
    and
    \[\frac{dv}{dx}=\frac{1}{x}\]
  5. Therefore,
    \[\frac{dy}{dx}=x^3\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x^3)\]
    \[=x^3\left(\frac{1}{x}\right)+\log x(3x^2)\]
    \[=x^2+3x^2\log x\]
  6. Hence,
    \[\boxed{\frac{dy}{dx}=x^2+3x^2\log x}\]
  7. Now differentiate the first derivative again with respect to \(x\):
    \[\frac{d^2y}{dx^2}=\frac{d}{dx}\left(x^2+3x^2\log x\right)\]
  8. Differentiate the first term:
    \[\frac{d}{dx}(x^2)=2x\]
  9. For the second term, \(3\) is a constant, so
    \[\frac{d}{dx}\left(3x^2\log x\right)=3\frac{d}{dx}\left(x^2\log x\right)\]
  10. Now \(x^2\log x\) is also a product. Applying the product rule:
    \[\frac{d}{dx}\left(x^2\log x\right)=x^2\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x^2)\]
    \[=x^2\left(\frac{1}{x}\right)+\log x(2x)\]
    \[=x+2x\log x\]
  11. Therefore,
    \[\frac{d}{dx}\left(3x^2\log x\right)=3\left(x+2x\log x\right)\]
    \[=3x+6x\log x\]
  12. Hence,
    \[\frac{d^2y}{dx^2}=2x+3x+6x\log x\]
  13. Combining the like terms \(2x\) and \(3x\):
    \[\frac{d^2y}{dx^2}=5x+6x\log x\]
  14. Taking \(x\) common:
    \[\boxed{\frac{d^2y}{dx^2}=x(5+6\log x)}\]
🎯 Exam Significance
Exam Significance
›

This question combines two important differentiation techniques: the product rule and higher order differentiation. It is particularly useful for understanding how a differentiation rule may need to be applied repeatedly. In a board examination, students should clearly show the first derivative before proceeding to the second derivative and should carefully differentiate the product \(x^2\log x\) in the second step.

Significance for Competitive Entrance Examination Aspirants

Functions involving powers of \(x\) multiplied by logarithmic functions occur frequently in advanced calculus. The ability to differentiate expressions such as \(x^n\log x\) efficiently is useful in problems involving maxima and minima, higher derivatives, Taylor expansions, approximation, and differential equations. Recognising the product structure immediately helps reduce calculation errors and saves time in objective-type examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Use the product rule when two functions are multiplied.

  2. For \(y=x^3\log x\), the first derivative is \(x^2+3x^2\log x\).

  3. The product rule is required again while differentiating \(x^2\log x\).

  4. The standard derivative \(\dfrac{d}{dx}(\log x)=\dfrac{1}{x}\) is used at both stages.

  5. After simplification, the second derivative is \(5x+6x\log x\).

  6. The factored form is \(x(5+6\log x)\).

  7. For the real-valued logarithm, \(x>0\).
← Q4
5 / 17  ·  29%
Q6 →
Q6
NUMERIC3 marks

Find the second order derivative of

\[ e^x\sin 5x \]
📘 Concept & Theory
Concept/Theory
›

The given function is the product of two functions, \(e^x\) and \(\sin 5x\). Therefore, the product rule is used to find the first derivative. The resulting expression is then differentiated again to obtain the second order derivative.

The product rule is

\[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\]

We also use the chain rule for the trigonometric functions:

\[ \frac{d}{dx}(\sin 5x)=5\cos 5x \]
\[ \frac{d}{dx}(\cos 5x)=-5\sin 5x \]

and the standard exponential derivative

\[ \frac{d}{dx}(e^x)=e^x. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(y=e^x\sin 5x\).

  2. Apply the product rule to obtain the first derivative.

  3. Use the chain rule while differentiating \(\sin 5x\).

  4. Differentiate the first derivative again.

  5. Apply the product rule separately to \(e^x\sin 5x\) and \(5e^x\cos 5x\).

  6. Use the chain rule for \(\cos 5x\).

  7. Combine like terms and factor the final expression.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Let
    \[y=e^x\sin 5x\]
  2. Since \(y\) is the product of \(e^x\) and \(\sin 5x\), apply the product rule:
    \[\frac{dy}{dx}=e^x\frac{d}{dx}(\sin 5x)+\sin 5x\frac{d}{dx}(e^x)\]
  3. Using
    \[\frac{d}{dx}(\sin 5x)=5\cos 5x\]
    and
    \[\frac{d}{dx}(e^x)=e^x\]
  4. we get
    \[\frac{dy}{dx}=e^x(5\cos 5x)+\sin 5x(e^x)\]
    \[\frac{dy}{dx}=e^x\sin 5x+5e^x\cos 5x\]
  5. Therefore,
    \[\boxed{\frac{dy}{dx}=e^x\sin 5x+5e^x\cos 5x}\]
  6. Now differentiate the first derivative again:
    \[\frac{d^2y}{dx^2}=\frac{d}{dx}\left(e^x\sin 5x+5e^x\cos 5x\right)\]
  7. Differentiate \(e^x\sin 5x\)
  8. Using the product rule,
    \[\frac{d}{dx}(e^x\sin 5x)=e^x\frac{d}{dx}(\sin 5x)+\sin 5x\frac{d}{dx}(e^x)\]
    \[=e^x(5\cos 5x)+e^x\sin 5x\]
    \[=e^x\sin 5x+5e^x\cos 5x\]
  9. Differentiate \(5e^x\cos 5x\)
  10. Since \(5\) is a constant,
    \[\frac{d}{dx}(5e^x\cos 5x)=5\frac{d}{dx}(e^x\cos 5x)\]
  11. Applying the product rule:
    \[\frac{d}{dx}(e^x\cos 5x)=e^x\frac{d}{dx}(\cos 5x)+\cos 5x\frac{d}{dx}(e^x)\]
  12. Using
    \[\frac{d}{dx}(\cos 5x)=-5\sin 5x\]
    and
    \[\frac{d}{dx}(e^x)=e^x\]
  13. we get
    \[\frac{d}{dx}(e^x\cos 5x)=e^x(-5\sin 5x)+e^x\cos 5x\]
    \[=e^x\cos 5x-5e^x\sin 5x\]
  14. Therefore,
    \[\frac{d}{dx}(5e^x\cos 5x)=5\left(e^x\cos 5x-5e^x\sin 5x\right)\]
    \[=5e^x\cos 5x-25e^x\sin 5x\]
  15. Add the two derivatives
  16. \[\frac{d^2y}{dx^2}=\left(e^x\sin 5x+5e^x\cos 5x\right)+\left(5e^x\cos 5x-25e^x\sin 5x\right)\]
  17. Group the \(\sin 5x\) and \(\cos 5x\) terms:
    \[\frac{d^2y}{dx^2}=e^x\sin 5x-25e^x\sin 5x+5e^x\cos 5x+5e^x\cos 5x\]
    \[=-24e^x\sin 5x+10e^x\cos 5x\]
  18. Taking \(e^x\) common:
    \[\frac{d^2y}{dx^2}=e^x\left(10\cos 5x-24\sin 5x\right)\]
  19. Taking \(2\) common:
    \[\boxed{\frac{d^2y}{dx^2}=2e^x\left(5\cos 5x-12\sin 5x\right)}\]
🎯 Exam Significance
Exam Significance
›

This problem combines three important differentiation techniques: the product rule, the chain rule, and higher order differentiation. It is particularly valuable because the product rule has to be applied again when finding the second derivative. Students should carefully retain the factor \(5\) generated by differentiating \(5x\). Missing this chain-rule factor changes the final answer.

Significance for Competitive Entrance Examination Aspirants

Expressions of the form \(e^{ax}\sin bx\) and \(e^{ax}\cos bx\) occur frequently in higher-order differentiation and differential equations. Efficient handling of the product rule and chain rule is useful for solving repeated-derivative problems quickly. Recognising the structure of exponential-trigonometric products also helps in advanced calculus and differential-equation problems.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Use the product rule for products of functions.

  2. Use the chain rule for \(\sin 5x\) and \(\cos 5x\).

  3. \(\dfrac{d}{dx}(\sin 5x)=5\cos 5x\).

  4. \(\dfrac{d}{dx}(\cos 5x)=-5\sin 5x\).

  5. \(\dfrac{d}{dx}(e^x)=e^x\).

  6. The second derivative requires differentiating the entire first derivative.

  7. The final result is \(\boxed{2e^x(5\cos 5x-12\sin 5x)}\).
← Q5
6 / 17  ·  35%
Q7 →
Q7
NUMERIC3 marks

Find the second order derivative of

\[ e^{6x}\cos 3x \]
📘 Concept & Theory
Concept/Theory
›

The given function is a product of an exponential function and a trigonometric function. Therefore, the product rule is required at both stages of differentiation.

The product rule is

\[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\]

Since the exponential and trigonometric functions contain \(6x\) and \(3x\), respectively, the chain rule is also required.

The required derivatives are

\[ \frac{d}{dx}(e^{6x})=6e^{6x} \]
\[ \frac{d}{dx}(\cos 3x)=-3\sin 3x \]
\[ \frac{d}{dx}(\sin 3x)=3\cos 3x. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(y=e^{6x}\cos 3x\).

  2. Apply the product rule to find \(\dfrac{dy}{dx}\).

  3. Use the chain rule to differentiate \(e^{6x}\) and \(\cos 3x\).

  4. Factor the first derivative to make the second differentiation easier.

  5. Differentiate the factored first derivative using the product rule.

  6. Differentiate \(2\cos 3x-\sin 3x\) carefully using the chain rule.

  7. Combine like terms and simplify the second derivative.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Let
    \[y=e^{6x}\cos 3x\]
  2. Since this is a product, use the product rule:
    \[\frac{dy}{dx}=e^{6x}\frac{d}{dx}(\cos 3x)+\cos 3x\frac{d}{dx}(e^{6x})\]
  3. Using the chain rule,
    \[\frac{d}{dx}(\cos 3x)=-\sin 3x\cdot\frac{d}{dx}(3x)=-3\sin 3x\]
    and
    \[\frac{d}{dx}(e^{6x})=e^{6x}\frac{d}{dx}(6x)=6e^{6x}\]
  4. Therefore,
    \[\frac{dy}{dx}=e^{6x}(-3\sin 3x)+\cos 3x(6e^{6x})\]
    \[=-3e^{6x}\sin 3x+6e^{6x}\cos 3x\]
  5. Taking \(3e^{6x}\) common:
    \[\boxed{\frac{dy}{dx}=3e^{6x}\left(2\cos 3x-\sin 3x\right)}\]
  6. Finding the Second Derivative
  7. Differentiate the first derivative again:
    \[\frac{d^2y}{dx^2}=3\frac{d}{dx}\left[e^{6x}\left(2\cos 3x-\sin 3x\right)\right]\]
  8. Here, \(3\) is a constant. Apply the product rule to the expression inside the brackets:
    \[\frac{d^2y}{dx^2}=3\left[e^{6x}\frac{d}{dx}\left(2\cos 3x-\sin 3x\right)+\left(2\cos 3x-\sin 3x\right)\frac{d}{dx}(e^{6x})\right]\]
  9. We already know that
    \[\frac{d}{dx}(e^{6x})=6e^{6x}\]
  10. Now differentiate the trigonometric expression:
    \[\frac{d}{dx}\left(2\cos 3x-\sin 3x\right)=2\frac{d}{dx}(\cos 3x)-\frac{d}{dx}(\sin 3x)\]
    \[=2(-3\sin 3x)-3\cos 3x\]
    \[=-6\sin 3x-3\cos 3x\]
  11. Substituting these derivatives:
    \[\frac{d^2y}{dx^2}=3\left[e^{6x}(-6\sin 3x-3\cos 3x)+\left(2\cos 3x-\sin 3x\right)(6e^{6x})\right]\]
  12. Take \(e^{6x}\) common:
    \[\frac{d^2y}{dx^2}=3e^{6x}\left[-6\sin 3x-3\cos 3x+6(2\cos 3x-\sin 3x)\right]\]
  13. Expand the bracket:
    \[\frac{d^2y}{dx^2}=3e^{6x}\left[-6\sin 3x-3\cos 3x+12\cos 3x-6\sin 3x\right]\]
  14. Combine like terms:
    \[-3\cos 3x+12\cos 3x=9\cos 3x\]
    and
    \[-6\sin 3x-6\sin 3x=-12\sin 3x\]
  15. Therefore,
    \[\frac{d^2y}{dx^2}=3e^{6x}\left(9\cos 3x-12\sin 3x\right)\]
  16. Taking \(3\) common from the bracket:
    \[\boxed{\frac{d^2y}{dx^2}=9e^{6x}\left(3\cos 3x-4\sin 3x\right)}\]
🎯 Exam Significance
Exam Significance
›

This problem is an important application of the product rule, chain rule, and second order differentiation. The coefficients \(6\) and \(3\) generated by the chain rule must be retained carefully. In a board examination, writing the first derivative separately and then differentiating it systematically helps avoid sign and coefficient errors and makes the solution easy to verify.

Significance for Competitive Entrance Examination Aspirants

Exponential-trigonometric functions such as \(e^{ax}\cos bx\) and \(e^{ax}\sin bx\) are important in higher calculus and differential equations. Repeated differentiation of these functions follows a useful structural pattern. Mastering the product rule and chain rule here helps in solving higher-order derivative questions efficiently and prepares students for problems involving linear differential equations with constant coefficients.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The product rule is required because \(e^{6x}\) and \(\cos 3x\) are multiplied.

  2. The chain rule is required because the arguments are \(6x\) and \(3x\).

  3. \(\dfrac{d}{dx}(e^{6x})=6e^{6x}\).

  4. \(\dfrac{d}{dx}(\cos 3x)=-3\sin 3x\).

  5. \(\dfrac{d}{dx}(\sin 3x)=3\cos 3x\).

  6. The first derivative is \(3e^{6x}(2\cos 3x-\sin 3x)\).

  7. The second derivative is \(9e^{6x}(3\cos 3x-4\sin 3x)\).
← Q6
7 / 17  ·  41%
Q8 →
Q8
NUMERIC3 marks

Find the second order derivative of

\[ \tan^{-1}x \]
📘 Concept & Theory
Concept/Theory
›

To find the second order derivative, first find the first derivative and then differentiate the resulting expression once again.

We use the standard derivative

\[ \frac{d}{dx}\left(\tan^{-1}x\right)=\frac{1}{1+x^2}. \]

The first derivative can be written as

\[ \frac{1}{1+x^2}=(1+x^2)^{-1}. \]
This form makes the chain rule straightforward to apply while finding the second derivative.

For a composite function \(u^n\), the chain rule gives

\[ \frac{d}{dx}(u^n)=nu^{n-1}\frac{du}{dx}. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(y=\tan^{-1}x\).

  2. Find the first derivative using the standard derivative of \(\tan^{-1}x\).

  3. Rewrite \(\dfrac{1}{1+x^2}\) as \((1+x^2)^{-1}\).

  4. Apply the chain rule to differentiate \((1+x^2)^{-1}\).

  5. Differentiate the inner function \(1+x^2\).

  6. Simplify the resulting expression to obtain the second derivative.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Let
    \[y=\tan^{-1}x\]
  2. Differentiate both sides with respect to \(x\):
    \[\frac{dy}{dx}=\frac{d}{dx}\left(\tan^{-1}x\right)\]
  3. Using the standard derivative
    \[\frac{d}{dx}\left(\tan^{-1}x\right)=\frac{1}{1+x^2},\]
  4. we obtain
    \[\boxed{\frac{dy}{dx}=\frac{1}{1+x^2}}\]
  5. Now differentiate the first derivative again:
    \[\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{1}{1+x^2}\right)\]
  6. Rewrite the fraction using a negative exponent:
    \[\frac{d^2y}{dx^2}=\frac{d}{dx}\left(1+x^2\right)^{-1}\]
  7. Apply the chain rule:
    \[\frac{d}{dx}\left(1+x^2\right)^{-1}=-1\left(1+x^2\right)^{-2}\frac{d}{dx}(1+x^2)\]
  8. Now differentiate the inner function:
    \[\frac{d}{dx}(1+x^2)=0+2x=2x\]
  9. Therefore,
    \[\frac{d^2y}{dx^2}=-1\left(1+x^2\right)^{-2}(2x)\]
    \[=-2x\left(1+x^2\right)^{-2}\]
  10. Using
    \[\left(1+x^2\right)^{-2}=\frac{1}{(1+x^2)^2},\]
  11. we get
    \[\boxed{\frac{d^2y}{dx^2}=-\frac{2x}{(1+x^2)^2}}\]
🎯 Exam Significance
Exam Significance
›

This question tests the standard derivative of an inverse trigonometric function followed by the application of the chain rule. It is important to recognise that the second differentiation is not simply another standard inverse-trigonometric derivative. After the first differentiation, the expression becomes \((1+x^2)^{-1}\), which must be differentiated using the chain rule. Showing the inner derivative \(2x\) explicitly makes the solution complete and reduces sign errors.

Significance for Competitive Entrance Examination Aspirants

Higher derivatives of inverse trigonometric functions appear in calculus problems involving Taylor expansions, concavity, maxima and minima, approximation, and differential equations. This example is also useful for developing speed in differentiating composite algebraic expressions. Recognising the structure \((1+x^2)^{-1}\) immediately after the first derivative allows the second derivative to be obtained efficiently.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. \(\dfrac{d}{dx}(\tan^{-1}x)=\dfrac{1}{1+x^2}\).

  2. The expression \(\dfrac{1}{1+x^2}\) can be rewritten as \((1+x^2)^{-1}\).

  3. The second differentiation requires the chain rule.

  4. The derivative of the inner function \(1+x^2\) is \(2x\).

  5. The negative sign comes from differentiating the power \(-1\).

  6. The second derivative is \(\boxed{-\dfrac{2x}{(1+x^2)^2}}\).
← Q7
8 / 17  ·  47%
Q9 →
Q9
NUMERIC3 marks

Find the second order derivative of

\[ \log(\log x) \]
📘 Concept & Theory
Concept/Theory
›

The given function is a logarithm of another logarithmic function. Therefore, the chain rule is required to find the first derivative. For the second derivative, the resulting expression is a product of two functions, so the product rule can be used.

We use the standard derivative

\[ \frac{d}{dx}(\log x)=\frac{1}{x}, \qquad x>0. \]

For the real-valued function \(\log(\log x)\), we additionally require

\[ \log x>0, \]
which gives
\[ x>1. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(u=\log x\), so that \(y=\log u\).

  2. Use the chain rule to find \(\dfrac{dy}{dx}\).

  3. Rewrite the first derivative as \(\dfrac{1}{x\log x}\).

  4. Differentiate the product \(\dfrac{1}{\log x}\cdot\dfrac{1}{x}\) using the product rule.

  5. Use the chain rule while differentiating \((\log x)^{-1}\).

  6. Combine the resulting fractions and simplify.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Let
    \[y=\log(\log x)\]
  2. Put
    \[u=\log x\]
  3. Then
    \[y=\log u\]
  4. Using the chain rule,
    \[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
  5. Now,
    \[\frac{dy}{du}=\frac{1}{u}\]
    and
    \[\frac{du}{dx}=\frac{d}{dx}(\log x)=\frac{1}{x}\]
  6. Therefore,
    \[\frac{dy}{dx}=\frac{1}{u}\cdot\frac{1}{x}\]
  7. Since \(u=\log x\),
    \[\boxed{\frac{dy}{dx}=\frac{1}{x\log x}}\]
  8. Finding the Second Derivative
  9. Rewrite the first derivative as
    \[\frac{dy}{dx}=\frac{1}{\log x}\cdot\frac{1}{x}.\]
  10. Differentiate both factors using the product rule:
    \[\frac{d^2y}{dx^2}=\frac{1}{x}\frac{d}{dx}\left(\frac{1}{\log x}\right)+\frac{1}{\log x}\frac{d}{dx}\left(\frac{1}{x}\right)\]
  11. Rewrite the first factor as a negative power:
    \[\frac{1}{\log x}=(\log x)^{-1}\]
  12. Using the chain rule,
    \[\frac{d}{dx}(\log x)^{-1}=-1(\log x)^{-2}\frac{d}{dx}(\log x)\]
    \[=-(\log x)^{-2}\cdot\frac{1}{x}\]
    \[=-\frac{1}{x(\log x)^2}\]
  13. Also,
    \[\frac{d}{dx}\left(\frac{1}{x}\right)=\frac{d}{dx}(x^{-1})=-x^{-2}=-\frac{1}{x^2}\]
  14. Substituting these results into the product-rule expression:
    \[\frac{d^2y}{dx^2}=\frac{1}{x}\left(-\frac{1}{x(\log x)^2}\right)+\frac{1}{\log x}\left(-\frac{1}{x^2}\right)\]
    \[=-\frac{1}{x^2(\log x)^2}-\frac{1}{x^2\log x}\]
  15. Take \(-\dfrac{1}{x^2}\) common:
    \[\frac{d^2y}{dx^2}=-\frac{1}{x^2}\left[\frac{1}{(\log x)^2}+\frac{1}{\log x}\right]\]
  16. Take the common denominator \((\log x)^2\):
    \[\frac{d^2y}{dx^2}=-\frac{1}{x^2}\left[\frac{1+\log x}{(\log x)^2}\right]\]
    \[=-\frac{1+\log x}{x^2(\log x)^2}\]
  17. Since
    \[x^2(\log x)^2=(x\log x)^2\]
  18. we obtain
    \[\boxed{\frac{d^2y}{dx^2}=-\frac{1+\log x}{(x\log x)^2}}\]
🎯 Exam Significance
Exam Significance
›

This problem is important because it combines the chain rule, product rule, and higher order differentiation. The first derivative requires the chain rule because \(\log x\) is itself the argument of another logarithm. The second derivative then requires careful differentiation of a product of reciprocal functions. Writing each intermediate step makes the handling of logarithmic terms and negative signs much clearer.

Significance for Competitive Entrance Examination Aspirants

Nested logarithmic functions are useful practice for problems involving composite functions and repeated differentiation. Such techniques appear in higher calculus, Taylor expansions, approximation, maxima and minima, and differential equations. The ability to switch between reciprocal and negative-power forms can make these calculations faster while reducing algebraic errors.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. For \(\log(\log x)\), use the chain rule for the first derivative.

  2. The first derivative is \(\dfrac{1}{x\log x}\).

  3. The second derivative can be found by applying the product rule to \(\dfrac{1}{\log x}\cdot\dfrac{1}{x}\).

  4. The derivative of \((\log x)^{-1}\) requires the chain rule.

  5. For the real-valued function \(\log(\log x)\), the domain is \(x>1\).

  6. The second derivative is

    \[ \boxed{ -\frac{1+\log x}{(x\log x)^2} }. \]
← Q8
9 / 17  ·  53%
Q10 →
Q10
NUMERIC3 marks

Find the second order derivative of

\[ \sin(\log x) \]
📘 Concept & Theory
Concept/Theory
›

The given function is a composite function because the argument of the sine function is \(\log x\). Therefore, the chain rule is required to find the first derivative.

After finding the first derivative, we obtain a quotient or product involving

\[ \frac{1}{x}\cos(\log x). \]
For the second derivative, we can use either the quotient rule or the product rule. Here, the quotient rule provides a direct route.

The required standard derivatives are

\[ \frac{d}{dx}(\sin u)=\cos u\frac{du}{dx} \]
\[ \frac{d}{dx}(\cos u)=-\sin u\frac{du}{dx} \]
and
\[ \frac{d}{dx}(\log x)=\frac{1}{x}. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(u=\log x\) and rewrite the function as \(y=\sin u\).

  2. Use the chain rule to find \(\dfrac{dy}{dx}\).

  3. Rewrite the first derivative as \(\dfrac{\cos(\log x)}{x}\).

  4. Apply the quotient rule to find the second derivative.

  5. Use the chain rule while differentiating \(\cos(\log x)\).

  6. Simplify the numerator and write the final result in compact form.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Let
    \[y=\sin(\log x)\]
  2. Put
    \[u=\log x\]
  3. Then
    \[y=\sin u\]
  4. Using the chain rule,
    \[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
  5. Now,
    \[\frac{dy}{du}=\cos u\]
    and
    \[\frac{du}{dx}=\frac{d}{dx}(\log x)=\frac{1}{x}\]
  6. Therefore,
    \[\frac{dy}{dx}=\cos u\cdot\frac{1}{x}\]
  7. Substituting \(u=\log x\),
    \[\boxed{\frac{dy}{dx}=\frac{\cos(\log x)}{x}}\]
  8. Finding the Second Derivative
  9. Differentiate the first derivative:
    \[\frac{d^2y}{dx^2}=\frac{d}{dx}\left[\frac{\cos(\log x)}{x}\right]\]
  10. Using the quotient rule
    \[\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\frac{df}{dx}-f\frac{dg}{dx}}{g^2},\]
  11. take
    \[f=\cos(\log x),\quad g=x\]
  12. First, differentiate \(f=\cos(\log x)\) using the chain rule:
    \[\frac{df}{dx}=-\sin(\log x)\frac{d}{dx}(\log x)\]
    \[=-\sin(\log x)\cdot\frac{1}{x}\]
    \[=-\frac{\sin(\log x)}{x}\]
  13. Also,
    \[\frac{dg}{dx}=\frac{d}{dx}(x)=1\]
  14. Substituting into the quotient rule:
    \[\frac{d^2y}{dx^2}=\frac{x\left(-\frac{\sin(\log x)}{x}\right)-\cos(\log x)(1)}{x^2}\]
  15. Cancel \(x\) in the first term:
    \[\frac{d^2y}{dx^2}=\frac{-\sin(\log x)-\cos(\log x)}{x^2}\]
  16. Therefore,
    \[\boxed{\frac{d^2y}{dx^2}=-\frac{\sin(\log x)+\cos(\log x)}{x^2}}\]
🎯 Exam Significance
Exam Significance
›

This question is an important application of the chain rule and higher order differentiation. The inner function \(\log x\) must be differentiated whenever the chain rule is applied. In the second derivative, students must also differentiate the factor \(\dfrac{1}{x}\). Showing these steps separately helps prevent the common error of differentiating only the trigonometric part.

Significance for Competitive Entrance Examination Aspirants

Composite functions involving trigonometric and logarithmic functions are common in advanced calculus. This problem develops the ability to recognise nested functions and differentiate them systematically. The same techniques are useful in questions involving higher derivatives, Taylor expansions, maxima and minima, approximation, and differential equations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. The function \(\sin(\log x)\) is a composite function.

  2. The chain rule is required for the first derivative.

  3. The first derivative is \(\dfrac{\cos(\log x)}{x}\).

  4. The second derivative can be found using either the quotient rule or product rule.

  5. The chain rule is required again while differentiating \(\cos(\log x)\).

  6. The final result is

    \[ \boxed{ \frac{d^2y}{dx^2} = -\frac{\sin(\log x)+\cos(\log x)}{x^2} }. \]
← Q9
10 / 17  ·  59%
Q11 →
Q11
NUMERIC3 marks
If \(y=5\cos x-3\sin x\;\) prove that \(\frac{d^2y}{dx^2}+y=0\)
📘 Concept & Theory
Concept/Theory
›

This is a verification problem involving a second order differential equation. We are given a function \(y\) and need to prove that it satisfies the differential equation

\[ \frac{d^2y}{dx^2}+y=0. \]

The procedure is straightforward: find the first derivative, differentiate once more to obtain the second derivative, substitute \(y\) and \(\dfrac{d^2y}{dx^2}\) into the left-hand side of the required equation, and simplify.

The relevant standard derivatives are

\[ \frac{d}{dx}(\cos x)=-\sin x \]
\[ \frac{d}{dx}(\sin x)=\cos x. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given function \(y=5\cos x-3\sin x\).

  2. Differentiate once to obtain \(\dfrac{dy}{dx}\).

  3. Differentiate the first derivative again to obtain \(\dfrac{d^2y}{dx^2}\).

  4. Substitute \(y\) and \(\dfrac{d^2y}{dx^2}\) into \(\dfrac{d^2y}{dx^2}+y\).

  5. Simplify the expression and show that it is equal to zero.

  6. Hence, the required differential equation is verified.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Given — \(y=5\cos x-3\sin x\)
  2. Find the first derivative
  3. Differentiate both sides with respect to \(x\):
    \[\frac{dy}{dx}=\frac{d}{dx}(5\cos x-3\sin x)\]
  4. Since \(5\) and \(3\) are constants,
    \[\frac{dy}{dx}=5\frac{d}{dx}(\cos x)-3\frac{d}{dx}(\sin x)\]
  5. Using
    \[\frac{d}{dx}(\cos x)=-\sin x\]
    and
    \[\frac{d}{dx}(\sin x)=\cos x\]
  6. we get
    \[\frac{dy}{dx}=5(-\sin x)-3(\cos x)\]
    \[\boxed{\frac{dy}{dx}=-5\sin x-3\cos x}\]
  7. Find the second derivative
  8. Differentiate the first derivative again:
    \[\frac{d^2y}{dx^2}=\frac{d}{dx}(-5\sin x-3\cos x)\]
    \[=-5\frac{d}{dx}(\sin x)-3\frac{d}{dx}(\cos x)\]
  9. Using the standard derivatives,
    \[=-5(\cos x)-3(-\sin x)\]
    \[=-5\cos x+3\sin x\]
  10. Therefore,
    \[\boxed{\frac{d^2y}{dx^2}=-5\cos x+3\sin x}\]
  12. To Prove — \(\frac{d^2y}{dx^2}+y=0\)
  13. Substitute the values of \(\dfrac{d^2y}{dx^2}\) and \(y\):
    \[\frac{d^2y}{dx^2}+y=(-5\cos x+3\sin x)+(5\cos x-3\sin x)\]
  14. Remove the brackets:
    \[=-5\cos x+3\sin x+5\cos x-3\sin x\]
  15. Group like terms:
    \[=(-5\cos x+5\cos x)+(3\sin x-3\sin x)\]
    \[=0+0\]
    \[\boxed{\frac{d^2y}{dx^2}+y=0}\]
  16. Hence Proved
🎯 Exam Significance
Exam Significance
›

This question tests the ability to calculate higher order derivatives and verify a given differential equation. It is important to show both derivatives clearly and then substitute them into the required expression. The cancellation of corresponding sine and cosine terms provides the final verification. Such questions are also useful for reinforcing the signs associated with derivatives of \(\sin x\) and \(\cos x\).

Significance for Competitive Entrance Examination Aspirants

Functions of the form \(A\cos x+B\sin x\) have a particularly useful property: differentiating twice reproduces the negative of the original function. Recognising this pattern can significantly reduce the work in higher derivative and differential equation problems. It also provides a foundation for understanding solutions of second order linear differential equations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. \(\dfrac{d}{dx}(\sin x)=\cos x\).

  2. \(\dfrac{d}{dx}(\cos x)=-\sin x\).

  3. The first derivative is \(-5\sin x-3\cos x\).

  4. The second derivative is \(-5\cos x+3\sin x\).

  5. The second derivative is the negative of the original function:

    \[ \frac{d^2y}{dx^2}=-y. \]

  6. Therefore,

    \[ \boxed{\frac{d^2y}{dx^2}+y=0}. \]

  7. The function \(5\cos x-3\sin x\) is a solution of the second order differential equation \(y''+y=0\).
← Q10
11 / 17  ·  65%
Q12 →
Q12
NUMERIC3 marks
If $y=\cos^{-1}x\;$ find $\frac{d^2y}{dx^2}\;$ in terms of \(y\) alone.
📘 Concept & Theory
Concept/Theory
›

Since the question specifically asks for the second derivative in terms of \(y\) alone, we should avoid replacing the final result with an expression containing \(x\). The most convenient approach is to first rewrite the inverse trigonometric relation as

\[ x=\cos y \]
and then differentiate implicitly with respect to \(x\).

From

\[ x=\cos y, \]
differentiating with respect to \(x\) requires the chain rule because \(y\) is a function of \(x\):
\[ \frac{d}{dx}(\cos y) = -\sin y\frac{dy}{dx}. \]

After obtaining \(\dfrac{dy}{dx}\), differentiate it again with respect to \(x\). The resulting expression will naturally contain only \(y\), which satisfies the requirement of the question.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Start with \(y=\cos^{-1}x\).

  2. Convert the inverse relation into \(x=\cos y\).

  3. Differentiate implicitly to find \(\dfrac{dy}{dx}\) in terms of \(y\).

  4. Rewrite the first derivative as \(-(\sin y)^{-1}\).

  5. Differentiate again using the chain rule.

  6. Substitute the first derivative into the second derivative.

  7. Simplify the result to obtain an expression involving \(y\) alone.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Given —
    \[y=\cos^{-1}x\]
  2. Taking cosine on both sides,
    \[\cos y=x\]
    or
    \[x=\cos y\]
  3. Find the First Derivative
  4. Differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}(x)=\frac{d}{dx}(\cos y)\]
  5. Since \(y\) is a function of \(x\), apply the chain rule:
    \[1=-\sin y\frac{dy}{dx}\]
  6. Therefore,
    \[\frac{dy}{dx}=-\frac{1}{\sin y}\]
    or
    \[ \boxed{ \frac{dy}{dx}=-(\sin y)^{-1} } \]
  7. Find the Second Derivative
  8. Differentiate the first derivative again with respect to \(x\):
    \[\frac{d^2y}{dx^2}=\frac{d}{dx}\left[-(\sin y)^{-1}\right]\]
  9. Take the negative sign outside:
    \[\frac{d^2y}{dx^2}=-\frac{d}{dx}\left[(\sin y)^{-1}\right]\]
  10. Using the chain rule,
    \[\frac{d}{dx}\left[(\sin y)^{-1}\right]=-(\sin y)^{-2}\frac{d}{dx}(\sin y)\]
  11. Again applying the chain rule,
    \[\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}\]
  12. Therefore,
    \[\frac{d}{dx}\left[(\sin y)^{-1}\right]=-(\sin y)^{-2}\cos y\frac{dy}{dx}\]
  13. Hence,
    \[\frac{d^2y}{dx^2}=-\left[-(\sin y)^{-2}\cos y\frac{dy}{dx}\right]\]
    \[=\frac{\cos y}{(\sin y)^2}\frac{dy}{dx}\]
  14. From the first derivative,
    \[\frac{dy}{dx}=-\frac{1}{\sin y}\]
  15. Substituting this value:
    \[\frac{d^2y}{dx^2}=\frac{\cos y}{(\sin y)^2}\left(-\frac{1}{\sin y}\right)\]
    \[=-\frac{\cos y}{(\sin y)^3}\]
  16. Therefore,
    \[\boxed{\frac{d^2y}{dx^2}=-\frac{\cos y}{\sin^3 y}}\]
  17. Using trigonometric notation,
    \[\frac{\cos y}{\sin^3 y}=\cot y\csc^2 y\]
  18. Hence, the required result is
    \[\boxed{\frac{d^2y}{dx^2}=-\cot y\csc^2 y}\]
🎯 Exam Significance
Exam Significance
›

This question is important because it tests implicit differentiation of an inverse trigonometric relation. The key requirement is that the final answer must be expressed in terms of \(y\) alone. Writing \(x=\cos y\) at the beginning makes this requirement much easier to satisfy. Students should also carefully apply the chain rule because \(y\) itself depends on \(x\).

Significance for Competitive Entrance Examination Aspirants

Expressing derivatives entirely in terms of the dependent variable is a useful technique in advanced calculus. It appears in problems involving inverse functions, parametric relationships, differential equations, and higher order derivatives. The relation \(x=\cos y\) also provides a systematic alternative to differentiating \(\cos^{-1}x\) directly.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. For \(y=\cos^{-1}x\), write the equivalent relation \(x=\cos y\).

  2. Implicit differentiation gives

    \[ \frac{dy}{dx}=-\frac{1}{\sin y}. \]

  3. The second differentiation requires the chain rule because \(y\) depends on \(x\).

  4. The second derivative is

    \[ -\frac{\cos y}{\sin^3y}. \]

  5. Using trigonometric identities,

    \[ \frac{\cos y}{\sin^3y} = \cot y\csc^2y. \]

  6. Therefore,

    \[ \boxed{ \frac{d^2y}{dx^2} = -\cot y\csc^2y }. \]
← Q11
12 / 17  ·  71%
Q13 →
Q13
NUMERIC3 marks
If \[ y=3\cos(\log x)+4\sin(\log x), \] show that \[ x^2y_2+xy_1+y=0, \] where \[ y_1=\frac{dy}{dx},\qquad y_2=\frac{d^2y}{dx^2}.\]
📘 Concept & Theory
Concept/Theory
›

This question uses the first and second derivatives of a function involving \(\log x\). Since

\[ \frac{d}{dx}(\log x)=\frac{1}{x}, \]
the chain rule is required when differentiating \(\sin(\log x)\) and \(\cos(\log x)\).

We must calculate \(y_1\) and \(y_2\), substitute them into \(x^2y_2+xy_1+y\), and simplify. The expression should reduce to zero.

For real-valued \(\log x\), we take \(x>0\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given function \(y\).

  2. Differentiate once to obtain \(y_1=\dfrac{dy}{dx}\).

  3. Differentiate \(y_1\) again to obtain \(y_2=\dfrac{d^2y}{dx^2}\).

  4. Substitute \(y\), \(y_1\), and \(y_2\) into \(x^2y_2+xy_1+y\).

  5. Collect the \(\cos(\log x)\) and \(\sin(\log x)\) terms.

  6. Show that the expression is equal to zero.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  23 steps
  1. Given — $y=3\cos(\log x)+4\sin(\log x)$
  2. Find \(y_1=\dfrac{dy}{dx}\)
  3. Differentiate both terms using the chain rule:
    \[ \begin{aligned} y_1 &=\frac{d}{dx}\left[3\cos(\log x)+4\sin(\log x)\right]\\ &=3\frac{d}{dx}\left[\cos(\log x)\right] +4\frac{d}{dx}\left[\sin(\log x)\right] \end{aligned} \]
  4. Now,
    \[\frac{d}{dx}\cos(\log x)=-\sin(\log x)\cdot\frac{1}{x}=-\frac{\sin(\log x)}{x}\]
    and
    \[\frac{d}{dx}\sin(\log x)=\cos(\log x)\cdot\frac{1}{x}=\frac{\cos(\log x)}{x}\]
  5. Therefore,
    \[\boxed{y_1=\frac{-3\sin(\log x)+4\cos(\log x)}{x}}\]
  6. Find \(y_2=\dfrac{d^2y}{dx^2}\)
  7. Differentiate \(y_1\):
    \[y_2=\frac{d}{dx}\left[\frac{-3\sin(\log x)+4\cos(\log x)}{x}\right]\]
  8. Separate the two terms:
    \[y_2=-3\frac{d}{dx}\left[\frac{\sin(\log x)}{x}\right]+4\frac{d}{dx}\left[\frac{\cos(\log x)}{x}\right]\]
  9. Differentiate \(\dfrac{\sin(\log x)}{x}\)
  10. Using the quotient rule,
    \[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\]
  11. Here,
    \[u=\sin(\log x),\quad v=x\]
  12. Hence,
    \[u'=\frac{\cos(\log x)}{x},\quad v'=1\]
  13. Therefore,
    \[\begin{aligned}\frac{d}{dx}\left[\frac{\sin(\log x)}{x}\right]&=\frac{x\left(\frac{\cos(\log x)}{x}\right)-\sin(\log x)(1)}{x^2}\\ &=\frac{\cos(\log x)-\sin(\log x)}{x^2}\end{aligned}\]
  14. Thus,
    \[-3\frac{d}{dx}\left[\frac{\sin(\log x)}{x}\right]=\frac{-3\cos(\log x)+3\sin(\log x)}{x^2}\]
  15. Differentiate \(\dfrac{\cos(\log x)}{x}\)
  16. Again, using the quotient rule,
    \[u=\cos(\log x),\qquad v=x\]
  17. Therefore,
    \[u'=-\frac{\sin(\log x)}{x},\quad v'=1\]
  18. Hence,
    \[\begin{aligned}\frac{d}{dx}\left[\frac{\cos(\log x)}{x}\right]&=\frac{x\left(-\frac{\sin(\log x)}{x}\right)-\cos(\log x)(1)}{x^2}\\ &=\frac{-\sin(\log x)-\cos(\log x)}{x^2}\end{aligned}\]
  19. Therefore,
    \[4\frac{d}{dx}\left[\frac{\cos(\log x)}{x}\right]=\frac{-4\sin(\log x)-4\cos(\log x)}{x^2}\]
  20. Combine both terms
  21. \[\begin{aligned}y_2&=\frac{-3\cos(\log x)+3\sin(\log x)}{x^2}+\frac{-4\sin(\log x)-4\cos(\log x)}{x^2}\\ &=\frac{-3\cos(\log x)+3\sin(\log x)-4\sin(\log x)-4\cos(\log x)}{x^2}\\ &=\frac{-7\cos(\log x)-\sin(\log x)}{x^2}\end{aligned}\]
  22. Thus,
    \[\boxed{y_2=-\frac{7\cos(\log x)+\sin(\log x)}{x^2}}\]
  23. Evaluate \(x^2y_2+xy_1+y\)
  24. We have
    \[y=3\cos(\log x)+4\sin(\log x),\]
    \[y_1=\frac{-3\sin(\log x)+4\cos(\log x)}{x},\]
    \[y_2=\frac{-7\cos(\log x)-\sin(\log x)}{x^2}\]
  25. Substituting these values,
    \[\begin{aligned}x^2y_2+xy_1+y&=x^2\left[\frac{-7\cos(\log x)-\sin(\log x)}{x^2}\right]\\ &\quad +x\left[\frac{-3\sin(\log x)+4\cos(\log x)}{x}\right]\\&\quad +\left[3\cos(\log x)+4\sin(\log x)\right]\end{aligned}\]
  26. Cancel \(x^2\) in the first term and \(x\) in the second term:
    \[\begin{aligned}x^2y_2+xy_1+y&=-7\cos(\log x)-\sin(\log x)\\ &\quad-3\sin(\log x)+4\cos(\log x)\\ &\quad+3\cos(\log x)+4\sin(\log x)\end{aligned}\]
  27. Collect the cosine terms:
    \[-7\cos(\log x)+4\cos(\log x)+3\cos(\log x)=0\]
  28. Collect the sine terms:
    \[-\sin(\log x)-3\sin(\log x)+4\sin(\log x)=0\]
  29. Therefore,
    \[\begin{aligned}x^2y_2+xy_1+y&=0+0\\&=0\end{aligned}\]
  30. Hence,
    \[\boxed{x^2y_2+xy_1+y=0}\]
  31. Hence Proved
🎯 Exam Significance
Exam Significance
›
  • This problem tests repeated differentiation involving \(\log x\).
  • It reinforces the chain rule and quotient rule.
  • It tests the ability to substitute first and second derivatives into a differential equation.
  • For Board examinations, clearly calculating \(y_1\) and \(y_2\) is important because intermediate steps carry marks.
  • For competitive examinations, recognizing the structure of the resulting differential equation can reduce lengthy calculations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. \[\frac{d}{dx}\cos(\log x)=-\frac{\sin(\log x)}{x}\]

  2. \[\frac{d}{dx}\sin(\log x)=\frac{\cos(\log x)}{x}\]

  3. \[y_1=\frac{-3\sin(\log x)+4\cos(\log x)}{x}\]

  4. \[y_2=\frac{-7\cos(\log x)-\sin(\log x)}{x^2}\]

  5. \[\boxed{x^2y_2+xy_1+y=0}\]
← Q12
13 / 17  ·  76%
Q14 →
Q14
NUMERIC3 marks
If \[ y=Ae^{mx}+Be^{nx}, \] show that \[ \frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny=0. \]
📘 Concept & Theory
Concept/Theory
›

This problem uses repeated differentiation of exponential functions and substitution into a given differential equation.

The key differentiation rule is

\[ \frac{d}{dx}\left(e^{kx}\right)=ke^{kx}, \]
where \(k\) is a constant.

Therefore,

\[ \frac{d}{dx}\left(Ae^{mx}\right)=Am e^{mx} \]
and
\[ \frac{d}{dx}\left(Be^{nx}\right)=Bn e^{nx}. \]

We need to find the first and second derivatives and then substitute them into

\[ \frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny. \]
The terms will cancel pairwise, giving zero.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given function \(y\).

  2. Differentiate once to obtain \(\dfrac{dy}{dx}\).

  3. Differentiate again to obtain \(\dfrac{d^2y}{dx^2}\).

  4. Substitute all three expressions into the required equation.

  5. Expand \((m+n)\dfrac{dy}{dx}\) carefully.

  6. Group the terms and show that they cancel to zero.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Given —
    \[y=Ae^{mx}+Be^{nx}\]
  2. Find the first derivative
  3. Differentiating both sides with respect to \(x\),
    \[\begin{aligned}\frac{dy}{dx}&=\frac{d}{dx}\left(Ae^{mx}+Be^{nx}\right)\\ &=A\frac{d}{dx}\left(e^{mx}\right)+B\frac{d}{dx}\left(e^{nx}\right)\end{aligned}\]
  4. Using the chain rule,
    \[\frac{d}{dx}\left(e^{mx}\right)=me^{mx}\]
    and
    \[\frac{d}{dx}\left(e^{nx}\right)=ne^{nx}\]
  5. Therefore,
    \[\boxed{\frac{dy}{dx}=Am e^{mx}+Bn e^{nx}}\]
  6. Find the second derivative
  7. Differentiate the first derivative again:
    \[\begin{aligned}\frac{d^2y}{dx^2}&=\frac{d}{dx}\left(Am e^{mx}+Bn e^{nx}\right)\\ &=Am\frac{d}{dx}\left(e^{mx}\right)+Bn\frac{d}{dx}\left(e^{nx}\right)\end{aligned}\]
  8. Again, applying the chain rule,
    \[\begin{aligned}\frac{d^2y}{dx^2}&=Am(me^{mx})+Bn(ne^{nx})\\ &=Am^2e^{mx}+Bn^2e^{nx}\end{aligned}\]
  9. Hence,
    \[\boxed{\frac{d^2y}{dx^2}=Am^2e^{mx}+Bn^2e^{nx}}\]
  11. Consider
    \[\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny\]
  12. Substituting the expressions obtained above,
    \[\begin{aligned}&\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny\\ &=\left(Am^2e^{mx}+Bn^2e^{nx}\right)\\ &\quad -(m+n)\left(Am e^{mx}+Bn e^{nx}\right)\\ &\quad +mn\left(Ae^{mx}+Be^{nx}\right)\end{aligned}\]
  13. Expand the middle term
    \[\begin{aligned}&(m+n)\left(Am e^{mx}+Bn e^{nx}\right)\\ &=Am(m+n)e^{mx}+Bn(m+n)e^{nx}\\ &=Am^2e^{mx}+Amne^{mx}+Bmne^{nx}+Bn^2e^{nx}\end{aligned}\]
  14. Also,
    \[mn\left(Ae^{mx}+Be^{nx}\right)=Amne^{mx}+Bmne^{nx}\]
  15. Substitute the expanded terms
    \[\begin{aligned}&\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny\\ &=Am^2e^{mx}+Bn^2e^{nx}\\ &\quad -\left(Am^2e^{mx}+Amne^{mx}+Bmne^{nx}+Bn^2e^{nx}\right)\\ &\quad +Amne^{mx}+Bmne^{nx}\end{aligned}\]
  16. Remove the brackets:
    \[\begin{aligned}&\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny\\ &=Am^2e^{mx}+Bn^2e^{nx}-Am^2e^{mx}-Amne^{mx}-Bmne^{nx}-Bn^2e^{nx}\\ &\quad +Amne^{mx}+Bmne^{nx}\end{aligned}\]
  17. Cancel like terms
  18. The terms cancel in pairs:
    \[Am^2e^{mx}-Am^2e^{mx}=0,\]
    \[Bn^2e^{nx}-Bn^2e^{nx}=0,\]
    \[-Amne^{mx}+Amne^{mx}=0,\]
    and
    \[-Bmne^{nx}+Bmne^{nx}=0.\]
  19. Therefore,
    \[\begin{aligned}\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny&=0\end{aligned}\]
  20. Hence Proved
🎯 Exam Significance
Exam Significance
›
  • This question tests repeated differentiation of exponential functions.
  • It reinforces the correct use of the chain rule for \(e^{mx}\) and \(e^{nx}\).
  • For Board examinations, showing the calculation of both derivatives and the cancellation of terms provides a complete proof.
  • For competitive examinations, recognizing the characteristic polynomial
    \[ r^2-(m+n)r+mn=(r-m)(r-n) \]
    provides a faster conceptual approach.
  • The result is an important connection between higher-order derivatives and linear differential equations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. \[\frac{d}{dx}(e^{mx})=me^{mx}\]

  2. \[\frac{d}{dx}(e^{nx})=ne^{nx}\]

  3. \[\frac{dy}{dx}=Am e^{mx}+Bn e^{nx}\]

  4. \[\frac{d^2y}{dx^2}=Am^2e^{mx}+Bn^2e^{nx}\]

  5. The associated characteristic polynomial is

    \[r^2-(m+n)r+mn=(r-m)(r-n).\]

  6. Therefore,

    \[\boxed{\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny=0} \]
← Q13
14 / 17  ·  82%
Q15 →
Q15
NUMERIC3 marks
If \[ y=500e^{7x}+600e^{-7x}, \] show that \[ \frac{d^2y}{dx^2}=49y. \]
📘 Concept & Theory
Concept/Theory
›

This question uses repeated differentiation of exponential functions. The important differentiation rules are

\[ \frac{d}{dx}(e^{ax})=ae^{ax} \]

and

\[ \frac{d}{dx}(e^{-ax})=-ae^{-ax}. \]

Notice that when \(e^{-7x}\) is differentiated twice, the negative sign appears twice and therefore becomes positive:

\[ \frac{d^2}{dx^2}(e^{-7x})=49e^{-7x}. \]

We will find \(y'\), then \(y''\), and finally compare \(y''\) with \(49y\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given function \(y\).

  2. Differentiate once to obtain \(y'\).

  3. Differentiate \(y'\) again to obtain \(y''\).

  4. Factor out \(49\) from \(y''\).

  5. Recognise the remaining expression as \(y\).

  6. Conclude that \(y''=49y\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Given — $ y=500e^{7x}+600e^{-7x}$
  2. Find the first derivative
  3. Differentiate both sides with respect to \(x\):
    \[\begin{aligned}y'&=\frac{d}{dx}\left(500e^{7x}+600e^{-7x}\right)\\ &=500\frac{d}{dx}\left(e^{7x}\right)+600\frac{d}{dx}\left(e^{-7x}\right)\end{aligned}\]
  4. Using
    \[\frac{d}{dx}(e^{7x})=7e^{7x}\]
    and
    \[\frac{d}{dx}(e^{-7x})=-7e^{-7x}\]
  5. we get
    \[\begin{aligned}y'&=500(7e^{7x})+600(-7e^{-7x})\\ &=3500e^{7x}-4200e^{-7x}\end{aligned}\]
  6. Therefore,
    \[\boxed{y'=3500e^{7x}-4200e^{-7x}}\]
  7. Find the second derivative
  8. Differentiate \(y'\) again:
    \[\begin{aligned}y''&=\frac{d}{dx}\left(3500e^{7x}-4200e^{-7x}\right)\\ &=3500\frac{d}{dx}\left(e^{7x}\right)-4200\frac{d}{dx}\left(e^{-7x}\right)\end{aligned}\]
  9. Applying the chain rule,
    \[\begin{aligned}y''&=3500(7e^{7x})-4200(-7e^{-7x})\\ &=24500e^{7x}+29400e^{-7x}\end{aligned}\]
  10. Factor out \(49\):
    \[\begin{aligned}y''&=49\left(500e^{7x}+600e^{-7x}\right)\end{aligned}\]
  11. But from the given equation,
    \[500e^{7x}+600e^{-7x}=y\]
  12. Therefore,
    \[\boxed{y''=49y}\]
  13. Hence Proved
🎯 Exam Significance
Exam Significance
›
  • This problem tests repeated differentiation of exponential functions.
  • It specifically tests the chain rule when the exponent is \(-7x\).
  • For Board examinations, writing both \(y'\) and \(y''\) clearly makes the verification complete.
  • For competitive examinations, recognising that both \(e^{7x}\) and \(e^{-7x}\) reproduce themselves after two differentiations can make the solution much faster.
  • The result is also a simple example of a second-order differential equation:
    \[ y''-49y=0. \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. \[ \frac{d}{dx}(e^{7x})=7e^{7x} \]

  2. \[ \frac{d}{dx}(e^{-7x})=-7e^{-7x} \]

  3. \[ \frac{d^2}{dx^2}(e^{7x})=49e^{7x} \]

  4. \[ \frac{d^2}{dx^2}(e^{-7x})=49e^{-7x} \]

  5. Therefore,

    \[ \boxed{\frac{d^2y}{dx^2}=49y} \]
← Q14
15 / 17  ·  88%
Q16 →
Q16
NUMERIC3 marks
If \[ e^y(x+1)=1, \] show that \[ \frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2. \]
📘 Concept & Theory
Concept/Theory
›

This is an implicit differentiation problem because \(y\) is related to \(x\) through the equation

\[ e^y(x+1)=1. \]
Since \(y\) is a function of \(x\), differentiation of \(e^y\) requires the chain rule:

\[ \frac{d}{dx}(e^y)=e^y\frac{dy}{dx}. \]

We first find \(\dfrac{dy}{dx}\), then differentiate it once more to obtain \(\dfrac{d^2y}{dx^2}\). Finally, we express the second derivative in terms of the first derivative.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Differentiate the given equation implicitly with respect to \(x\).

  2. Use the product rule on \(e^y(x+1)\).

  3. Use the chain rule for \(\dfrac{d}{dx}(e^y)\).

  4. Simplify to obtain \(\dfrac{dy}{dx}\).

  5. Differentiate \(\dfrac{dy}{dx}\) to obtain \(\dfrac{d^2y}{dx^2}\).

  6. Compare the result with \(\left(\dfrac{dy}{dx}\right)^2\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  20 steps
  1. Given — $e^y(x+1)=1$
  2. Differentiate implicitly
  3. Differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}\left[e^y(x+1)\right]=\frac{d}{dx}(1)\]
  4. Since the left-hand side is a product of \(e^y\) and \(x+1\), apply the product rule:
    \[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}.\]
  5. Taking
    \[u=e^y,\quad v=x+1\]
  6. we get
    \[\frac{d}{dx}\left[e^y(x+1)\right]=e^y\frac{d}{dx}(x+1)+(x+1)\frac{d}{dx}(e^y)\]
  7. Now,
    \[\frac{d}{dx}(x+1)=1\]
  8. and, by the chain rule,
    \[\frac{d}{dx}(e^y)=e^y\frac{dy}{dx}\]
  9. Therefore,
    \[e^y+(x+1)e^y\frac{dy}{dx}=0\]
  10. Find \(\dfrac{dy}{dx}\)
  11. Rearranging,
    \[(x+1)e^y\frac{dy}{dx}=-e^y\]
  12. Since \(e^y\neq0\), divide both sides by \((x+1)e^y\):
    \[\frac{dy}{dx}=-\frac{e^y}{(x+1)e^y}\]
  13. Cancel \(e^y\):
    \[\boxed{\frac{dy}{dx}=-\frac{1}{x+1}}\]
  14. Find \(\dfrac{d^2y}{dx^2}\)
  15. Differentiate the first derivative again:
    \[\begin{aligned}\frac{d^2y}{dx^2}&=\frac{d}{dx}\left[-\frac{1}{x+1}\right]\\ &=-\frac{d}{dx}\left[(x+1)^{-1}\right]\end{aligned}\]
  16. Using the chain rule,
    \[\frac{d}{dx}(x+1)^{-1}=-1(x+1)^{-2}\frac{d}{dx}(x+1)\]
  17. Since
    \[\frac{d}{dx}(x+1)=1\]
  18. we obtain
    \[\frac{d}{dx}(x+1)^{-1}=-(x+1)^{-2}\]
  19. Hence,
    \[\begin{aligned}\frac{d^2y}{dx^2}&=-\left[-(x+1)^{-2}\right]\\ &=(x+1)^{-2}\\ &=\frac{1}{(x+1)^2}\end{aligned}\]
  20. Therefore,
    \[\boxed{\frac{d^2y}{dx^2}=\frac{1}{(x+1)^2}}\]
  21. Express the result in terms of \(\dfrac{dy}{dx}\)
  22. We already obtained
    \[\frac{dy}{dx}=-\frac{1}{x+1}\]
  23. Squaring both sides gives
    \[\left(\frac{dy}{dx}\right)^2=\left(-\frac{1}{x+1}\right)^2=\frac{1}{(x+1)^2}\]
  24. But
    \[\frac{d^2y}{dx^2}=\frac{1}{(x+1)^2}\]
  25. Therefore,
    \[\boxed{\frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2}\]
  26. Hence Proved
🎯 Exam Significance
Exam Significance
›
  • This problem is a standard application of implicit differentiation.
  • It tests the product rule when both factors involve functions of \(x\).
  • It tests the chain rule through the derivative
    \[ \frac{d}{dx}(e^y)=e^y\frac{dy}{dx}. \]
  • For Board examinations, writing the product-rule and chain-rule steps explicitly makes the solution complete and avoids sign errors.
  • For competitive examinations, recognising that the given equation immediately gives
    \[ e^y=\frac{1}{x+1} \]
    provides an alternative route for verification.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. For a function \(y=y(x)\),

    \[ \frac{d}{dx}(e^y)=e^y\frac{dy}{dx}. \]

  2. The first derivative is

    \[ \frac{dy}{dx}=-\frac{1}{x+1}. \]

  3. The second derivative is

    \[ \frac{d^2y}{dx^2}=\frac{1}{(x+1)^2}. \]

  4. Squaring the first derivative gives

    \[ \left(\frac{dy}{dx}\right)^2 = \frac{1}{(x+1)^2}. \]

  5. Therefore,

    \[ \boxed{ \frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2 }. \]
← Q15
16 / 17  ·  94%
Q17 →
Q17
NUMERIC3 marks
If \[ y=\left(\tan^{-1}x\right)^2, \] show that \[(1+x^2)y_2+2xy_1=\frac{2}{1+x^2}\]
📘 Concept & Theory
Concept/Theory
›

This problem requires repeated differentiation of

\[ y=(\tan^{-1}x)^2. \]
We use the chain rule for the first derivative and then differentiate again to obtain \(y_2\).

The standard derivative used here is

\[ \frac{d}{dx}\left(\tan^{-1}x\right)=\frac{1}{1+x^2}. \]

There appears to be a typographical error in the question as supplied. The identity that follows from the given function is

\[ \boxed{(x^2+1)y_2+2xy_1=2}. \]
The additional factor \((x^2+1)\) multiplying \(2xy_1\) in the supplied question makes the stated identity incorrect in general. The calculation below demonstrates this explicitly.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Put \(u=\tan^{-1}x\), so that \(y=u^2\).

  2. Find \(y_1=\dfrac{dy}{dx}\) using the chain rule.

  3. Differentiate \(y_1\) to obtain \(y_2=\dfrac{d^2y}{dx^2}\).

  4. Substitute \(y_1\) and \(y_2\) into the intended identity.

  5. Show that \((x^2+1)y_2+2xy_1=2\).

  6. Verify why the extra factor \((x^2+1)\) in the supplied expression cannot give \(2\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  18 steps
  1. Given — $y=(\tan^{-1}x)^2$
  2. Find \(y_1\)
  3. Using the chain rule,
    \[\begin{aligned}y_1&=\frac{d}{dx}\left[(\tan^{-1}x)^2\right]\\ &=2\tan^{-1}x\cdot\frac{d}{dx}(\tan^{-1}x)\\ &=2\tan^{-1}x\cdot\frac{1}{1+x^2}\end{aligned}\]
  4. Therefore,
    \[\boxed{y_1=\frac{2\tan^{-1}x}{1+x^2}}\]
  5. Find \(y_2\)
  6. Differentiate \(y_1\):
    \[y_2=\frac{d}{dx}\left[\frac{2\tan^{-1}x}{1+x^2}\right]\]
  7. Using the quotient rule,
    \[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\]
  8. Take
    \[u=2\tan^{-1}x,\quad v=1+x^2\]
  9. Then
    \[u'=\frac{2}{1+x^2}\]
    and
    \[v'=2x\]
  10. Therefore,
    \[\begin{aligned}y_2&=\frac{(1+x^2)\left(\frac{2}{1+x^2}\right)-(2\tan^{-1}x)(2x)}{(1+x^2)^2}\\ &=\frac{2-4x\tan^{-1}x}{(1+x^2)^2}\end{aligned}\]
  11. Taking \(2\) common,
    \[\boxed{y_2=\frac{2\left(1-2x\tan^{-1}x\right)}{(1+x^2)^2}}\]
  12. Verify the correct identity
  13. Consider
    \[(x^2+1)y_2+2xy_1\]
  14. Substituting \(y_2\) and \(y_1\),
    \[\begin{aligned}(x^2+1)y_2+2xy_1&=(x^2+1)\left[\frac{2(1-2x\tan^{-1}x)}{(1+x^2)^2}\right]\\ &\quad +2x\left[\frac{2\tan^{-1}x}{1+x^2}\right]\end{aligned}\]
  15. Since \(x^2+1=1+x^2\),
    \[\begin{aligned}(x^2+1)y_2+2xy_1&=\frac{2(1-2x\tan^{-1}x)}{1+x^2}+\frac{4x\tan^{-1}x}{1+x^2}\\ &=\frac{2-4x\tan^{-1}x+4x\tan^{-1}x}{1+x^2}\\ &=\frac{2}{1+x^2}\end{aligned}\]
  16. This reveals an important point: even the commonly expected form
    \[ (x^2+1)y_2+2xy_1=2 \]
    is not correct for the given function.
  17. The actual identity obtained from the derivatives is
    \[\boxed{(x^2+1)y_2+2xy_1=\frac{2}{1+x^2}}\]
  18. Derive the exact differential identity
  19. Since
    \[ y_1=\frac{2\tan^{-1}x}{1+x^2}, \]
  20. we can multiply by \(1+x^2\):
    \[(1+x^2)y_1=2\tan^{-1}x\]
  21. Differentiate both sides:
    \[\frac{d}{dx}\left[(1+x^2)y_1\right]=\frac{d}{dx}\left(2\tan^{-1}x\right)\]
  22. Using the product rule on the left-hand side,
    \[(1+x^2)y_2+2xy_1=\frac{2}{1+x^2}\]
  23. Thus the exact result is
    \[\boxed{(1+x^2)y_2+2xy_1=\frac{2}{1+x^2}}\]
🎯 Exam Significance
Exam Significance
›
  • This problem is an application of repeated differentiation of an inverse trigonometric function.
  • It reinforces the chain rule for
    \[ (\tan^{-1}x)^2. \]
  • The second derivative requires careful application of the quotient rule.
  • Differentiating
    \[ (1+x^2)y_1=2\tan^{-1}x \]
    provides a shorter route to the differential identity.
  • In an examination, the algebra should be checked against the stated identity rather than forcing the calculation to produce the expected answer.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. \[ y=(\tan^{-1}x)^2 \]

  2. \[ y_1=\frac{2\tan^{-1}x}{1+x^2} \]

  3. \[ y_2= \frac{2(1-2x\tan^{-1}x)} {(1+x^2)^2} \]

  4. The exact identity is

    \[ \boxed{ (1+x^2)y_2+2xy_1 = \frac{2}{1+x^2} }. \]

  5. The expression supplied in the question does not follow from the given function and should be checked against the source.

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NCERT Class 12 Maths Exercise 5.7 Solutions
NCERT Class 12 Maths Exercise 5.7 Solutions — Complete Notes & Solutions · academia-aeternum.com
Explore the NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability Exercise 5.7 solutions with clear, step-by-step explanations designed for CBSE Board and competitive examination preparation. This exercise focuses on higher-order derivatives and their applications, including second-order differentiation, exponential functions, logarithmic functions, inverse trigonometric functions, and implicit differentiation. Each solution explains the underlying concept, differentiation…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Frequently Asked Questions

    Exercise 5.7 of NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability focuses on second-order derivatives and applications of higher-order differentiation.

    A second-order derivative is the derivative of the first derivative. It is written as d²y/dx² or y2 and measures how the first derivative changes with respect to x.

    The chain rule, product rule, quotient rule, and standard derivatives of exponential, logarithmic, and inverse trigonometric functions are important for solving Exercise 5.7.

    First find the first derivative dy/dx and then differentiate the resulting expression once more with respect to x to obtain d²y/dx².

    The chain rule is required when differentiating composite functions such as sin(log x), cos(log x), (tan?¹x)², and exponential functions with non-constant-looking inner expressions.

    Implicit differentiation is used when x and y are related by an equation rather than y being explicitly given as a function of x. Both sides are differentiated with respect to x while treating y as a function of x.

    Students should learn the standard derivative formulas, practise first and second derivatives carefully, show all intermediate steps, and verify the required identity at the end.

    Yes. The exercise strengthens higher-order differentiation, chain rule, implicit differentiation, and algebraic simplification, which are useful for calculus-based competitive examination problems.

    Common mistakes include missing chain-rule factors, incorrect signs in exponential and trigonometric derivatives, errors while applying the quotient rule, and skipping simplification steps.

    Revise the standard derivatives first, then practise finding y1 and y2, followed by identity-based questions. Focus especially on sign changes and factors introduced by the chain rule.

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