Ch 5  ·  Q–
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Chapter 5 Exercise 5.6 Solutions

Continuity and Differentiability

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 5 Exercise 5.6

Class 12 Mathematics Exercise 5.6 NCERT Solutions Continuity and Differentiability Class 12 Mathematics Chapter 5 CBSE Board Exam JEE Main CUET Parametric Differentiation Differentiation Derivatives Chain Rule Product Rule Logarithmic Differentiation Trigonometric Differentiation
11 Questions
25–35 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
If \(x\) and \(y\) are connected parametrically by the given equations, without eliminating the parameter, find \(\frac{dy}{dx}\) \[ \begin{aligned} x &= 2at^2,\\ y &= at^4 \end{aligned} \]
📘 Concept & Theory
Concept/Theory
›

When both \(x\) and \(y\) are expressed in terms of a third variable \(t\), the variable \(t\) is called a parameter. Such equations are called parametric equations.

For example, if

\[ x=f(t), \quad y=g(t), \]
then \(x\) and \(y\) are connected through the parameter \(t\).

To differentiate \(y\) with respect to \(x\), there is no need to eliminate \(t\). We differentiate both \(x\) and \(y\) with respect to \(t\), and then use the chain rule:

\[ \boxed{ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} } \]

This formula is valid at points where

\[ \frac{dx}{dt}\neq 0. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given parametric equations for \(x\) and \(y\).

  2. Differentiate \(x\) with respect to the parameter \(t\) to obtain \(\dfrac{dx}{dt}\).

  3. Differentiate \(y\) with respect to \(t\) to obtain \(\dfrac{dy}{dt}\).

  4. Use the parametric differentiation formula

    \[ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}. \]

  5. Substitute the derivatives and simplify the result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  10 steps
  1. Write the given parametric equations.
    \[x=2at^2\]
    \[y=at^4\]
    Here, \(t\) is the parameter and \(a\) is treated as a constant.
  2. Differentiate \(x\) with respect to \(t\).
    \[x=2at^2\]
  3. Since \(2a\) is a constant, differentiating with respect to \(t\) gives
    \[\frac{dx}{dt}=2a\frac{d}{dt}(t^2)\]
    \[\frac{dx}{dt}=2a(2t)\]
    \[\boxed{\frac{dx}{dt}=4at}\]
  4. Differentiate \(y\) with respect to \(t\).
    \[y=at^4\]
  5. Since \(a\) is a constant,
    \[\frac{dy}{dt}=a\frac{d}{dt}(t^4)\]
    \[\frac{dy}{dt}=a(4t^3)\]
    \[\boxed{\frac{dy}{dt}=4at^3}\]
  6. Apply the formula for parametric differentiation.
    \[\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\]
  7. Substituting the values obtained above:
    \[\frac{dy}{dx}=\frac{4at^3}{4at}\]
  8. Simplify
    \[\frac{dy}{dx}=\frac{4at^3}{4at}\]
  9. For \(at\neq0\), cancel the common factor \(4at\):
    \[\frac{dy}{dx}=t^2\]
  10. Therefore,
    \[\boxed{\frac{dy}{dx}=t^2}\]
🎯 Exam Significance
Exam Significance
›

This question tests one of the fundamental applications of differentiation: differentiation of parametric equations. In board examinations, students are expected to remember the relation

\[ \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \]
and apply it accurately without unnecessarily eliminating the parameter.

The question also checks basic differentiation of powers and proper simplification of algebraic expressions. Writing the intermediate derivatives explicitly helps avoid errors involving powers of \(t\).

Significance for Competitive Entrance Exam Aspirants

Parametric differentiation is frequently useful in calculus problems involving curves represented by a parameter. Competitive examinations may combine this technique with higher derivatives, tangent and normal problems, increasing or decreasing functions, or evaluation at a particular parameter value.

A key speed-building idea is to recognise immediately that when

\[ x=f(t),\quad y=g(t), \]
the derivative is obtained through
\[ \frac{dy}{dx}=\frac{g'(t)}{f'(t)}. \]
However, the condition \(f'(t)\neq0\) should also be kept in mind.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. When \(x\) and \(y\) are given in terms of a parameter \(t\), use parametric differentiation.

  2. The fundamental formula is

    \[ \boxed{\frac{dy}{dx}=\frac{dy/dt}{dx/dt}}. \]

  3. Differentiate \(x\) and \(y\) separately with respect to \(t\).

  4. There is no need to eliminate the parameter when the question specifically asks for parametric differentiation.

  5. For the given equations,

    \[ x=2at^2,\quad y=at^4, \]
    we obtain
    \[ \boxed{\frac{dy}{dx}=t^2}. \]

  6. The formula requires

    \[ \frac{dx}{dt}\neq0 \]
    at the point under consideration.
↑ Top
1 / 11  ·  9%
Q2 →
Q2
NUMERIC3 marks
If \(x\) and \(y\) are connected parametrically by the given equations, without eliminating the parameter, find \(\frac{dy}{dx}\) \[ \begin{aligned} x &= a\cos\theta,\\ y &= b\cos\theta \end{aligned} \]
📘 Concept & Theory
Concept/Theory
›

When \(x\) and \(y\) are both expressed in terms of a parameter such as \(\theta\), we use parametric differentiation.

If

\[ x=f(\theta),\quad y=g(\theta), \]
then, provided
\[ \frac{dx}{d\theta}\neq0, \]
the derivative of \(y\) with respect to \(x\) is
\[ \boxed{ \frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}} {\dfrac{dx}{d\theta}} } \]

Equivalently, using the chain rule,

\[ \frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx}. \]

In this problem, \(a\) and \(b\) are constants, while \(\theta\) is the parameter. Therefore, \(a\) and \(b\) are treated as constants while differentiating with respect to \(\theta\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given parametric equations for \(x\) and \(y\).

  2. Differentiate \(x\) with respect to \(\theta\).

  3. Differentiate \(y\) with respect to \(\theta\).

  4. Use

    \[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}. \]

  5. Substitute the derivatives and simplify the result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  13 steps
  1. Write the given parametric equations.
    \[x=a\cos\theta\]
    \[y=b\cos\theta\]
    Here, \(\theta\) is the parameter, while \(a\) and \(b\) are constants.
  2. Differentiate \(x\) with respect to \(\theta\).
    \[x=a\cos\theta\]
  3. Since \(a\) is a constant,
    \[\frac{dx}{d\theta}=a\frac{d}{d\theta}(\cos\theta)\]
    \[\frac{dx}{d\theta}=a(-\sin\theta)\]
  4. Therefore,
    \[\boxed{\frac{dx}{d\theta}=-a\sin\theta}\]
  5. Differentiate \(y\) with respect to \(\theta\).
    \[y=b\cos\theta\]
  6. Since \(b\) is a constant,
    \[\frac{dy}{d\theta}=b\frac{d}{d\theta}(\cos\theta)\]
    \[\frac{dy}{d\theta}=b(-\sin\theta)\]
  7. Therefore,
    \[\boxed{\frac{dy}{d\theta}=-b\sin\theta}\]
  8. Apply the formula for parametric differentiation.
    \[\frac{dy}{dx}=\frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}\]
  9. Substituting the values obtained above,
    \[\frac{dy}{dx}=\frac{-b\sin\theta}{-a\sin\theta}\]
  10. Simplify
    \[\frac{dy}{dx}=\frac{b\sin\theta}{a\sin\theta}\]
  11. For
    \[\sin\theta\neq0,\]
  12. the common factor \(\sin\theta\) cancels:
    \[\frac{dy}{dx}=\frac{b}{a}\]
  13. Hence,
    \[ \boxed{ \frac{dy}{dx}=\frac{b}{a} } \]
🎯 Exam Significance
Exam Significance
›

This problem tests the standard formula for differentiation of parametric equations and the derivatives of basic trigonometric functions. It is important for board examinations because a small notation error, such as confusing \(\sin\theta\) with \(\sin^{-1}\theta\), can change the meaning of the entire solution.

A complete board-level solution should clearly show

\[ \frac{dx}{d\theta} \]
and
\[ \frac{dy}{d\theta} \]
before applying
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}. \]
This makes the method transparent and reduces the possibility of losing marks due to skipped steps.

Significance for Competitive Entrance Exam Aspirants

For competitive examinations, the main skill tested here is rapid recognition of the parametric differentiation pattern. When both parametric equations contain a common factor, the derivative may simplify substantially.

In this problem,

\[ \frac{dx}{d\theta}=-a\sin\theta \]
and
\[ \frac{dy}{d\theta}=-b\sin\theta. \]
Recognising the common factor \(-\sin\theta\) allows the expression to be simplified quickly to
\[ \frac{b}{a}. \]

Such cancellation-based reasoning is useful in objective questions where the final answer may be required without eliminating the parameter.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. For parametric equations \(x=f(\theta)\) and \(y=g(\theta)\), use

    \[ \boxed{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} }. \]

  2. The derivative of \(\cos\theta\) is

    \[ \frac{d}{d\theta}(\cos\theta)=-\sin\theta. \]

  3. Constants such as \(a\) and \(b\) remain unchanged while differentiating with respect to \(\theta\).

  4. Do not confuse \(\sin\theta\) with \(\sin^{-1}\theta\). The former is the sine function, whereas the latter denotes the inverse sine function.

  5. For the given equations,

    \[ x=a\cos\theta,\quad y=b\cos\theta, \]
    we obtain
    \[ \boxed{\frac{dy}{dx}=\frac{b}{a}}. \]

  6. The cancellation of \(\sin\theta\) is valid where

    \[ \sin\theta\neq0. \]

  7. The final result is independent of the parameter \(\theta\), indicating a constant slope.
← Q1
2 / 11  ·  18%
Q3 →
Q3
NUMERIC3 marks
If \(x\) and \(y\) are connected parametrically by the given equations, without eliminating the parameter, find \(\frac{dy}{dx}\) \[ \begin{aligned} x &= \sin t,\\ y &= \cos 2t \end{aligned} \]
📘 Concept & Theory
Concept/Theory
›

When \(x\) and \(y\) are expressed in terms of a third variable \(t\), the variable \(t\) is called a parameter. Such equations are called parametric equations.

To find the derivative of \(y\) with respect to \(x\), we do not need to eliminate the parameter. Instead, we differentiate both equations with respect to \(t\) and use the chain rule:

\[ \boxed{ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}} {\dfrac{dx}{dt}} } \]

Equivalently,

\[ \frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx}. \]

The formula is applicable at points where

\[ \frac{dx}{dt}\neq0. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Differentiate \(x=\sin t\) with respect to \(t\).

  2. Find \(\dfrac{dt}{dx}\), if using the chain-rule form.

  3. Differentiate \(y=\cos 2t\) with respect to \(t\), carefully applying the chain rule.

  4. Use

    \[ \frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx}. \]

  5. Use the double-angle identity

    \[ \sin 2t=2\sin t\cos t \]
    to simplify the result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  18 steps
  1. Differentiate \(x\) with respect to \(t\).
    \[x=\sin t\]
  2. Differentiating both sides with respect to \(t\),
    \[\frac{dx}{dt}=\frac{d}{dt}(\sin t)\]
  3. Therefore,
    \[\boxed{\frac{dx}{dt}=\cos t}\]
  4. Hence, where \(\cos t\neq0\),
    \[\frac{dt}{dx}=\frac{1}{\dfrac{dx}{dt}}\]
    \[\boxed{\frac{dt}{dx}=\frac{1}{\cos t}}\]
  5. Differentiate \(y\) with respect to \(t\).
    \[y=\cos 2t\]
  6. Differentiating with respect to \(t\), we must apply the chain rule because the angle is \(2t\).
    \[\frac{dy}{dt}=\frac{d}{dt}(\cos 2t)\]
  7. Using
    \[\frac{d}{dt}(\cos u)=-\sin u\frac{du}{dt},\]
    where
    \[u=2t,\]
  8. we have
    \[\frac{du}{dt}=2.\]
  9. Therefore,
    \[\frac{dy}{dt}=-\sin 2t(2)\]
  10. Hence,
    \[\boxed{\frac{dy}{dt}=-2\sin 2t}\]
  11. Apply the formula for parametric differentiation.
    \[\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}\]
  12. Substituting the values obtained above,
    \[\frac{dy}{dx}=(-2\sin 2t)\left(\frac{1}{\cos t}\right)\]
  13. Therefore,
    \[\frac{dy}{dx}=\frac{-2\sin 2t}{\cos t}\]
  14. Use the double-angle identity.
  15. Recall that
    \[\sin 2t=2\sin t\cos t.\]
  16. Substituting this into the expression,
    \[\frac{dy}{dx}=\frac{-2(2\sin t\cos t)}{\cos t}\]
  17. Multiplying the numerator,
    \[\frac{dy}{dx}=\frac{-4\sin t\cos t}{\cos t}\]
  18. For \(\cos t\neq0\), cancel \(\cos t\):
    \[\frac{dy}{dx}=-4\sin t\]
  19. Hence,
    \[\boxed{\frac{dy}{dx}=-4\sin t}\]
🎯 Exam Significance
Exam Significance
›

This problem combines three important Class 12 calculus skills:

  1. Parametric differentiation.
  2. Chain rule for composite trigonometric functions.
  3. Use of the double-angle identity
    \[ \sin 2t=2\sin t\cos t. \]

It is therefore a useful example for board preparation because it demonstrates that a complete solution requires not only the parametric differentiation formula but also correct application of the chain rule and trigonometric simplification.

Writing

\[ \frac{dx}{dt},\quad \frac{dy}{dt} \]
separately before calculating \(\dfrac{dy}{dx}\) makes the solution systematic and helps prevent sign and factor errors.

Significance for Competitive Entrance Exam Aspirants

For competitive examinations, this question illustrates an efficient pattern: differentiate both parametric equations, form their ratio, and simplify using a standard trigonometric identity.

The expression

\[ \frac{-2\sin 2t}{\cos t} \]
can be simplified rapidly by recognising
\[ \sin 2t=2\sin t\cos t. \]
This immediately produces
\[ -4\sin t. \]

The problem also reinforces the importance of checking the denominator. Since

\[ \frac{dx}{dt}=\cos t, \]
the standard parametric derivative formula requires
\[ \cos t\neq0. \]

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. For parametric equations \(x=f(t)\) and \(y=g(t)\),

    \[ \boxed{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} }. \]

  2. For

    \[ x=\sin t, \]
    we have
    \[ \frac{dx}{dt}=\cos t. \]

  3. When differentiating \(\cos 2t\), the chain rule must be applied:

    \[ \frac{d}{dt}(\cos 2t)=-2\sin 2t. \]

  4. The identity

    \[ \sin 2t=2\sin t\cos t \]
    simplifies the derivative.

  5. The parameter need not be eliminated.

  6. The required condition for the standard parametric derivative is

    \[ \cos t\neq0. \]

  7. The final derivative is

    \[ \boxed{\frac{dy}{dx}=-4\sin t}. \]
← Q2
3 / 11  ·  27%
Q4 →
Q4
NUMERIC3 marks
If \(x\) and \(y\) are connected parametrically by the given equations, without eliminating the parameter, find \(\frac{dy}{dx}\) \[ \begin{aligned} x &= 4t,\\ y &= \frac{4}{t} \end{aligned} \]
📘 Concept & Theory
Concept/Theory
›

When \(x\) and \(y\) are expressed in terms of a parameter \(t\), the derivative \(\dfrac{dy}{dx}\) can be found directly by differentiating both equations with respect to \(t\).

The fundamental formula for parametric differentiation is

\[ \boxed{ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}} {\dfrac{dx}{dt}} } \]

Alternatively, using the chain rule,

\[ \frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx}. \]

In this problem, \(t\) occurs in the denominator of \(y\), so it is useful to rewrite the reciprocal as a negative power:

\[ \frac{4}{t}=4t^{-1}. \]

We can then apply the power rule

\[ \frac{d}{dt}(t^n)=nt^{n-1}. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Differentiate \(x=4t\) with respect to \(t\).

  2. Find \(\dfrac{dt}{dx}\).

  3. Rewrite \(y=\dfrac{4}{t}\) as \(4t^{-1}\).

  4. Differentiate \(y\) with respect to \(t\) using the power rule.

  5. Use

    \[ \frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx}. \]

  6. Simplify the resulting expression.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Differentiate \(x\) with respect to \(t\).
  2. \[x=4t\]
  3. Differentiating both sides with respect to \(t\),
    \[\frac{dx}{dt}=\frac{d}{dt}(4t)\]
  4. Since \(4\) is a constant,
    \[\boxed{\frac{dx}{dt}=4}\]
  5. Therefore,
    \[\frac{dt}{dx}=\frac{1}{\dfrac{dx}{dt}}\]
  6. and hence
    \[\boxed{\frac{dt}{dx}=\frac{1}{4}}\]
  7. Differentiate \(y\) with respect to \(t\).
    \[y=\frac{4}{t}\]
  8. Rewrite the expression using a negative exponent:
    \[y=4t^{-1}\]
  9. Applying the power rule,
    \[ \frac{dy}{dt} = 4\frac{d}{dt}(t^{-1}) \]
  10. Using
    \[\frac{d}{dt}(t^n)=nt^{n-1},\]
  11. with \(n=-1\),
    \[\frac{dy}{dt}=4(-1)t^{-1-1}\]
    \[\frac{dy}{dt}=-4t^{-2}\]
  12. Since
    \[t^{-2}=\frac{1}{t^2}\]
  13. we obtain
    \[\boxed{\frac{dy}{dt}=-\frac{4}{t^2}}\]
  14. Apply the formula for parametric differentiation.
    \[\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}\]
  15. Substituting
    \[\frac{dy}{dt}=-\frac{4}{t^2}\]
    and
    \[\frac{dt}{dx}=\frac14\]
  16. we get
    \[\frac{dy}{dx}=\left(-\frac{4}{t^2}\right)\left(\frac14\right)\]
  17. Simplify
    \[\frac{dy}{dx}=-\frac{4}{4t^2}\]
  18. Therefore,
    \[\boxed{\frac{dy}{dx}=-\frac{1}{t^2}}\]
🎯 Exam Significance
Exam Significance
›

This question is a straightforward application of parametric differentiation, but it also tests the correct differentiation of a reciprocal expression. Students should be comfortable converting

\[ \frac{4}{t} \]
into
\[ 4t^{-1} \]
before applying the power rule.

A complete board-level solution should clearly show

\[ \frac{dx}{dt} \]
and
\[ \frac{dy}{dt} \]
before calculating \(\dfrac{dy}{dx}\). This avoids errors with negative powers and signs.

The restriction

\[ t\neq0 \]
is also important because the original expression \(\dfrac4t\) is undefined at \(t=0\).

Significance for Competitive Entrance Exam Aspirants

This problem reinforces a useful speed technique: whenever a parametric equation contains a reciprocal, immediately convert it into a negative power and apply the power rule.

The calculation can then be performed quickly:

\[ x=4t \quad\Rightarrow\quad \frac{dx}{dt}=4 \]
\[ y=4t^{-1} \quad\Rightarrow\quad \frac{dy}{dt}=-4t^{-2} \]
Hence,
\[ \frac{dy}{dx} = \frac{-4t^{-2}}{4} = -t^{-2} = -\frac1{t^2}. \]

This compact approach is particularly useful in objective-type questions where efficient algebraic manipulation saves time.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. For parametric equations, use

    \[ \boxed{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} }. \]

  2. A reciprocal can be rewritten as a negative power:

    \[ \frac{4}{t}=4t^{-1}. \]

  3. Using the power rule,

    \[ \frac{d}{dt}(4t^{-1}) = -4t^{-2} = -\frac4{t^2}. \]

  4. For \(x=4t\),

    \[ \frac{dx}{dt}=4. \]

  5. The parameter must satisfy

    \[ t\neq0. \]

  6. The final derivative is

    \[ \boxed{ \frac{dy}{dx}=-\frac1{t^2} }. \]
← Q3
4 / 11  ·  36%
Q5 →
Q5
NUMERIC3 marks
If \(x\) and \(y\) are connected parametrically by the given equations, without eliminating the parameter, find \(\frac{dy}{dx}\) \[ \begin{aligned} x &= \cos\theta-\cos2\theta,\\ y &= \sin\theta-\sin2\theta \end{aligned} \]
📘 Concept & Theory
Concept/Theory
›

When \(x\) and \(y\) are expressed in terms of a parameter \(\theta\), we can find \(\dfrac{dy}{dx}\) directly without eliminating the parameter.

The fundamental formula for parametric differentiation is

\[ \boxed{ \frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}} {\dfrac{dx}{d\theta}} } \]

Equivalently, using the chain rule,

\[ \frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx}. \]

Since the equations contain \(\cos2\theta\) and \(\sin2\theta\), the chain rule must be used carefully. In particular,

\[ \frac{d}{d\theta}(\cos2\theta) = -2\sin2\theta \]
and
\[ \frac{d}{d\theta}(\sin2\theta) = 2\cos2\theta. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Differentiate \(x\) with respect to \(\theta\).

  2. Differentiate \(y\) with respect to \(\theta\), applying the chain rule to \(2\theta\).

  3. Use

    \[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}. \]

  4. Substitute the derivatives into the formula.

  5. Simplify the resulting expression carefully without unnecessarily eliminating the parameter.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  18 steps
  1. Differentiate \(x\) with respect to \(\theta\).
  2. \[x=\cos\theta-\cos2\theta\]
  3. Differentiating both sides with respect to \(\theta\),
    \[ \frac{dx}{d\theta} = \frac{d}{d\theta}(\cos\theta) - \frac{d}{d\theta}(\cos2\theta) \]
  4. We know that
    \[\frac{d}{d\theta}(\cos\theta)=-\sin\theta\]
  5. For the second term, applying the chain rule,
    \[\frac{d}{d\theta}(\cos2\theta)=-\sin2\theta\cdot\frac{d}{d\theta}(2\theta)\]
    \[=-2\sin2\theta\]
  6. Therefore,
    \[\frac{dx}{d\theta}=-\sin\theta-(-2\sin2\theta)\]
    \[\boxed{\frac{dx}{d\theta}=-\sin\theta+2\sin2\theta}\]
  7. Hence, where
    \[\frac{dx}{d\theta}\neq0,\]
  8. we may write
    \[\frac{d\theta}{dx}=\frac{1}{\dfrac{dx}{d\theta}}\]
  9. Therefore,
    \[\boxed{\frac{d\theta}{dx}=\frac{1}{-\sin\theta+2\sin2\theta}}\]
  10. Differentiate \(y\) with respect to \(\theta\).
  11. \[y=\sin\theta-\sin2\theta\]
  12. Differentiating both sides,
    \[\frac{dy}{d\theta}=\frac{d}{d\theta}(\sin\theta)-\frac{d}{d\theta}(\sin2\theta)\]
  13. We have
    \[\frac{d}{d\theta}(\sin\theta)=\cos\theta\]
  14. Applying the chain rule to \(\sin2\theta\),
    \[\frac{d}{d\theta}(\sin2\theta)=\cos2\theta\cdot\frac{d}{d\theta}(2\theta)\]
    \[=2\cos2\theta\]
  15. Therefore,
    \[\frac{dy}{d\theta}=\cos\theta-2\cos2\theta\]
  16. Hence,
    \[\boxed{\frac{dy}{d\theta}=\cos\theta-2\cos2\theta}\]
  17. Apply the formula for parametric differentiation.
    \[\frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta}{dx}\]
  18. Substituting the values obtained above,
    \[\frac{dy}{dx}=\left(\cos\theta-2\cos2\theta\right)\left(\frac{1}{-\sin\theta+2\sin2\theta}\right)\]
  19. Therefore,
    \[\boxed{\frac{dy}{dx}=\frac{\cos\theta-2\cos2\theta}{-\sin\theta+2\sin2\theta}}\]
  20. Equivalently, writing the denominator in a more conventional order,
    \[ \boxed{ \frac{dy}{dx} = \frac{\cos\theta-2\cos2\theta} {2\sin2\theta-\sin\theta} } \]
🎯 Exam Significance
Exam Significance
›

This problem is particularly useful for board preparation because it combines parametric differentiation with the chain rule for trigonometric functions. Students must correctly handle the inner function \(2\theta\).

The most important intermediate results are

\[ \frac{dx}{d\theta} = -\sin\theta+2\sin2\theta \]
and
\[ \frac{dy}{d\theta} = \cos\theta-2\cos2\theta. \]

Showing these steps explicitly makes the solution easier to verify and helps prevent sign errors. It also demonstrates the proper application of the parametric differentiation formula expected in a written examination.

Significance for Competitive Entrance Exam Aspirants

This question reinforces a high-frequency calculus technique: differentiate parametrically first and simplify only after forming the ratio.

For speed-based problems, it is useful to recognise immediately that

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}. \]
There is no need to solve either equation for \(\theta\).

Competitive problems may further ask for values of the derivative at a specified parameter, horizontal or vertical tangents, or conditions involving the slope. Correctly obtaining \(\dfrac{dx}{d\theta}\) and \(\dfrac{dy}{d\theta}\) is therefore an important foundation.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. For parametric equations,

    \[\boxed{\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}}\]

  2. Always apply the chain rule when differentiating functions such as

    \[\cos2\theta\quad\text{and}\quad\sin2\theta\]

  3. The correct derivative of \(\cos2\theta\) is

    \[ -2\sin2\theta. \]

  4. The correct derivative of \(\sin2\theta\) is

    \[ 2\cos2\theta. \]

  5. For the given \(x\),

    \[ \frac{dx}{d\theta} = -\sin\theta+2\sin2\theta. \]

  6. For the given \(y\),

    \[ \frac{dy}{d\theta} = \cos\theta-2\cos2\theta. \]

  7. Do not incorrectly replace \(\cos2\theta\) by \(\cos\theta\).

  8. The parametric derivative is defined through the standard formula where

    \[ 2\sin2\theta-\sin\theta\neq0. \]
← Q4
5 / 11  ·  45%
Q6 →
Q6
NUMERIC3 marks
If \(x\) and \(y\) are connected parametrically by the given equations, without eliminating the parameter, find \(\frac{dy}{dx}\) \[ \begin{aligned} x &= a(\theta-\sin\theta),\\ y &= a(1+\cos\theta) \end{aligned} \]
📘 Concept & Theory
Concept/Theory
›

When \(x\) and \(y\) are expressed in terms of a parameter \(\theta\), the derivative \(\dfrac{dy}{dx}\) can be obtained without eliminating the parameter.

The fundamental formula for parametric differentiation is

\[ \boxed{ \frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}} {\dfrac{dx}{d\theta}} } \]

Equivalently, by the chain rule,

\[ \frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx}. \]

This problem also requires the use of the half-angle identities

\[ 1-\cos\theta = 2\sin^2\frac{\theta}{2} \]
and
\[ \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}. \]

These identities allow the final derivative to be simplified into a compact half-angle form.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Differentiate \(x=a(\theta-\sin\theta)\) with respect to \(\theta\).

  2. Differentiate \(y=a(1+\cos\theta)\) with respect to \(\theta\).

  3. Apply

    \[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}. \]

  4. Cancel the common constant \(a\).

  5. Use half-angle identities to simplify the resulting trigonometric expression.

  6. Obtain the final result in terms of \(\cot(\theta/2)\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  24 steps
  1. Differentiate \(x\) with respect to \(\theta\).
  2. \[x=a(\theta-\sin\theta)\]
  3. Expanding,
    \[x=a\theta-a\sin\theta\]
  4. Differentiating both sides with respect to \(\theta\),
    \[\frac{dx}{d\theta}=a\frac{d}{d\theta}(\theta)-a\frac{d}{d\theta}(\sin\theta)\]
  5. Since
    \[\frac{d}{d\theta}(\theta)=1\]
    and
    \[\frac{d}{d\theta}(\sin\theta)=\cos\theta,\]
  6. we obtain
    \[\frac{dx}{d\theta}=a-a\cos\theta\]
  7. Taking \(a\) common,
    \[\boxed{\frac{dx}{d\theta}=a(1-\cos\theta)}\]
  8. Hence, wherever
    \[a(1-\cos\theta)\neq0,\]
  9. we have
    \[\frac{d\theta}{dx}=\frac{1}{a(1-\cos\theta)}\]
  10. Therefore,
    \[\boxed{\frac{d\theta}{dx}=\frac{1}{a(1-\cos\theta)}}\]
  11. Differentiate \(y\) with respect to \(\theta\)
  12. \[y=a(1+\cos\theta)\]
  13. Expanding,
    \[y=a+a\cos\theta\]
  14. Differentiating with respect to \(\theta\),
    \[\frac{dy}{d\theta}=\frac{d}{d\theta}(a)+a\frac{d}{d\theta}(\cos\theta)\]
  15. Since \(a\) is a constant,
    \[\frac{d}{d\theta}(a)=0\]
    and
    \[\frac{d}{d\theta}(\cos\theta)=-\sin\theta\]
  16. Therefore,
    \[\frac{dy}{d\theta}=0-a\sin\theta\]
  17. Hence,
    \[\boxed{\frac{dy}{d\theta}=-a\sin\theta}\]
  18. Apply the parametric differentiation formula
    \[\frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta}{dx}\]
  19. Substituting the values obtained above,
    \[\frac{dy}{dx}=(-a\sin\theta)\left(\frac{1}{a(1-\cos\theta)}\right)\]
  20. Therefore,
    \[\frac{dy}{dx}=\frac{-a\sin\theta}{a(1-\cos\theta)}\]
  21. Cancelling the common non-zero factor \(a\),
    \[\boxed{\frac{dy}{dx}=-\frac{\sin\theta}{1-\cos\theta}}\]
  22. Apply the half-angle identities.
  23. Use
    \[\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\]
    and
    \[1-\cos\theta=2\sin^2\frac{\theta}{2}\]
  24. Substituting these into the derivative,
    \[\frac{dy}{dx}=-\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}\]
  25. Cancel the common factor \(2\):
    \[\frac{dy}{dx}=-\frac{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{\sin^2\frac{\theta}{2}}\]
  26. Cancel one factor of
    \[ \sin\frac{\theta}{2} \]
    from the numerator and denominator:
    \[\frac{dy}{dx}=-\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}\]
  27. Since
    \[\cot\frac{\theta}{2}=\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}},\]
  28. we obtain
    \[\boxed{\frac{dy}{dx}=-\cot\frac{\theta}{2}}\]
🎯 Exam Significance
Exam Significance
›

This problem is an important Class 12 example because it combines parametric differentiation with trigonometric identities. A complete board solution should demonstrate three distinct stages:

  1. Find \(\dfrac{dx}{d\theta}\) and \(\dfrac{dy}{d\theta}\).
  2. Form the ratio
    \[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}. \]
  3. Simplify the resulting trigonometric expression using half-angle identities.

The question is also useful for developing accuracy with signs. The derivative of \(\cos\theta\) is

\[ -\sin\theta, \]
so the derivative of
\[ a(1+\cos\theta) \]
is
\[ -a\sin\theta. \]

Significance for Competitive Entrance Exam Aspirants

This problem demonstrates an efficient combination of calculus and trigonometric manipulation. Once the parametric derivatives are obtained, the essential expression is

\[ -\frac{\sin\theta}{1-\cos\theta}. \]

Recognising the half-angle identities immediately transforms this into

\[ -\cot\frac{\theta}{2}. \]

This type of simplification is valuable in competitive examinations, especially when a subsequent question asks for the slope at a particular parameter value or requires the derivative to be expressed in a specific trigonometric form.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. For parametric equations,

    \[ \boxed{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} }. \]

  2. For

    \[ x=a(\theta-\sin\theta), \]
    we obtain
    \[ \frac{dx}{d\theta}=a(1-\cos\theta). \]

  3. For

    \[ y=a(1+\cos\theta), \]
    we obtain
    \[ \frac{dy}{d\theta}=-a\sin\theta. \]

  4. The common constant \(a\) cancels when forming \(\dfrac{dy}{dx}\).

  5. The half-angle identities

    \[ \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \]
    and
    \[ 1-\cos\theta = 2\sin^2\frac{\theta}{2} \]
    are essential for the final simplification.

  6. Do not confuse \(\sin\theta\) and \(\cos\theta\) with inverse trigonometric functions such as \(\sin^{-1}\theta\) and \(\cos^{-1}\theta\).

  7. The final derivative is

    \[ \boxed{ \frac{dy}{dx} = -\cot\frac{\theta}{2} }. \]
← Q5
6 / 11  ·  55%
Q7 →
Q7
NUMERIC3 marks
If \(x\) and \(y\) are connected parametrically by the given equations, without eliminating the parameter, find \(\frac{dy}{dx}\) \[ x=\frac{\sin^3t}{\sqrt{\cos 2t}}, \quad y=\frac{\cos^3t}{\sqrt{\cos 2t}} \]
📘 Concept & Theory
Concept/Theory
›

When \(x\) and \(y\) are given as functions of a common parameter \(t\), the derivative \(\frac{dy}{dx}\) can be found directly without eliminating the parameter by using parametric differentiation:

\[ \boxed{\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}} \]

Here, both \(x\) and \(y\) contain the common denominator \(\sqrt{\cos 2t}\). Therefore, the quotient rule and chain rule must be applied carefully.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Differentiate \(x\) with respect to \(t\).

  2. Differentiate \(y\) with respect to \(t\).

  3. Use the chain rule to differentiate \(\sqrt{\cos 2t}\).

  4. Apply \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\).

  5. Simplify the result using \(\sin 2t=2\sin t\cos t\) and \(\cos 2t=\cos^2t-\sin^2t\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  24 steps
  1. Differentiate \(x\)
  2. We have
    \[x=\frac{\sin^3t}{\sqrt{\cos 2t}}=\sin^3t(\cos 2t)^{-1/2}\]
  3. Using the product rule,
    \[\frac{dx}{dt}=\frac{d}{dt}(\sin^3t)(\cos 2t)^{-1/2}+\sin^3t\frac{d}{dt}(\cos 2t)^{-1/2}\]
  4. First,
    \[\frac{d}{dt}(\sin^3t)=3\sin^2t\cos t\]
  5. Now differentiate the second factor using the chain rule:
    \[\frac{d}{dt}(\cos 2t)^{-1/2}=-\frac12(\cos 2t)^{-3/2}\frac{d}{dt}(\cos 2t)\]
  6. Since
    \[\frac{d}{dt}(\cos 2t)=-2\sin 2t\]
  7. we obtain
    \[\frac{d}{dt}(\cos 2t)^{-1/2}=-\frac12(\cos 2t)^{-3/2}(-2\sin 2t)\]
    \[=\frac{\sin 2t}{(\cos 2t)^{3/2}}\]
  8. Therefore,
    \[\frac{dx}{dt}=\frac{3\sin^2t\cos t}{\sqrt{\cos 2t}}+\frac{\sin^3t\sin 2t}{(\cos 2t)^{3/2}}\]
  9. Taking \((\cos 2t)^{3/2}\) as the common denominator,
    \[\frac{dx}{dt}=\frac{3\sin^2t\cos t\cos 2t+\sin^3t\sin 2t}{(\cos 2t)^{3/2}}\]
  10. Factor out \(\sin^2t\):
    \[\boxed{\frac{dx}{dt}=\frac{\sin^2t\left(3\cos t\cos 2t+\sin t\sin 2t\right)}{(\cos 2t)^{3/2}}}\]
  11. Differentiate \(y\)
  12. We have
    \[y=\frac{\cos^3t}{\sqrt{\cos 2t}}=\cos^3t(\cos 2t)^{-1/2}\]
  13. Using the product rule,
    \[\frac{dy}{dt}=\frac{d}{dt}(\cos^3t)(\cos 2t)^{-1/2}+\cos^3t\frac{d}{dt}(\cos 2t)^{-1/2}\]
  14. First,
    \[\frac{d}{dt}(\cos^3t)=-3\cos^2t\sin t\]
  15. From the previous step, Hence,
    \[\frac{dy}{dt}=-\frac{3\cos^2t\sin t}{\sqrt{\cos 2t}}+\frac{\cos^3t\sin 2t}{(\cos 2t)^{3/2}}\]
  16. Taking \((\cos 2t)^{3/2}\) as the common denominator,
    \[\frac{dy}{dt}=\frac{-3\cos^2t\sin t\cos 2t+\cos^3t\sin 2t}{(\cos 2t)^{3/2}}\]
  17. Factor out \(\cos^2t\):
    \[\boxed{\frac{dy}{dt}=\frac{\cos^2t\left(-3\sin t\cos 2t+\cos t\sin 2t\right)}{(\cos 2t)^{3/2}}}\]
  18. Find \(\frac{dy}{dx}\)
  19. Using the parametric differentiation formula,
    \[\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\]
  20. Substituting the expressions obtained above,
    \[\frac{dy}{dx}=\frac{\dfrac{\cos^2t\left(-3\sin t\cos 2t+\cos t\sin 2t\right)}{(\cos 2t)^{3/2}}}{\dfrac{\sin^2t\left(3\cos t\cos 2t+\sin t\sin 2t\right)}{(\cos 2t)^{3/2}}}\]
  21. The common factor \((\cos 2t)^{3/2}\) cancels:
    \[\boxed{\frac{dy}{dx}=\frac{\cos^2t\left(-3\sin t\cos 2t+\cos t\sin 2t\right)}{\sin^2t\left(3\cos t\cos 2t+\sin t\sin 2t\right)}}\]
  22. Simplify the Result
  23. Use
    \[\sin 2t=2\sin t\cos t\]
  24. For the numerator bracket,
    \[-3\sin t\cos 2t+\cos t\sin 2t\]
  25. Substituting \(\cos 2t=\cos^2t-\sin^2t\) and \(\sin 2t=2\sin t\cos t\),
    \[=-3\sin t(\cos^2t-\sin^2t)+\cos t(2\sin t\cos t)\]
    \[=-3\sin t\cos^2t+3\sin^3t+2\sin t\cos^2t\]
    \[=\sin t(3\sin^2t-\cos^2t)\]
  26. For the denominator bracket,
    \[3\cos t\cos 2t+\sin t\sin 2t\]
    \[=3\cos t(\cos^2t-\sin^2t)+\sin t(2\sin t\cos t)\]
    \[=3\cos^3t-3\cos t\sin^2t+2\sin^2t\cos t\]
    \[=\cos t(3\cos^2t-\sin^2t)\]
  27. Therefore,
    \[\frac{dy}{dx}=\frac{\cos^2t\sin t(3\sin^2t-\cos^2t)}{\sin^2t\cos t(3\cos^2t-\sin^2t)}\]
  28. Cancel \(\sin t\) and \(\cos t\):
    \[\boxed{\frac{dy}{dx}=\frac{\cos t}{\sin t}\frac{3\sin^2t-\cos^2t}{3\cos^2t-\sin^2t}}\]
  29. Hence,
    \[\boxed{\frac{dy}{dx}=\cot t\,\frac{3\sin^2t-\cos^2t}{3\cos^2t-\sin^2t}}\]
🎯 Exam Significance
Exam Significance
›

This problem is important because it combines parametric differentiation with the quotient rule, product rule and chain rule. In a board examination, writing the derivatives of \(x\) and \(y\) separately and then applying \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\) provides a clear and logically complete solution. Correct handling of the derivative of \(\cos 2t\) is especially important because a missing factor of \(2\) or an incorrect sign changes the final answer.

Significance for Competitive Entrance Exams

For competitive examinations, the key skill is efficient simplification. After finding \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), recognising the identities \(\sin 2t=2\sin t\cos t\) and \(\cos 2t=\cos^2t-\sin^2t\) reduces the expression rapidly. This also strengthens algebraic manipulation and trigonometric simplification skills used in calculus-based entrance-examination problems.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. For parametric equations, use \(\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}\).

  2. Rewrite radicals as powers when it makes differentiation easier.

  3. Use the chain rule carefully for composite functions such as \(\cos 2t\).

  4. Remember:

    \[ \frac{d}{dt}(\cos 2t)^{-1/2} = \frac{\sin 2t}{(\cos 2t)^{3/2}}. \]

  5. Use double-angle identities only after obtaining the parametric derivative when they simplify the expression.

  6. Always check that the denominator in \(\frac{dy}{dx}\) is non-zero at the point under consideration.
← Q6
7 / 11  ·  64%
Q8 →
Q8
NUMERIC3 marks
If \(x\) and \(y\) are connected parametrically by the given equations, without eliminating the parameter, find \(\frac{dy}{dx}\) \[ \begin{aligned} x &= a\left(\cos t+\log\tan\frac{t}{2}\right),\\ y &= \sin t \end{aligned} \]
📘 Concept & Theory
Concept/Theory
›

When \(x\) and \(y\) are expressed in terms of a parameter \(t\), we can find \(\dfrac{dy}{dx}\) directly without eliminating the parameter.

The formula for parametric differentiation is

\[ \boxed{ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}} {\dfrac{dx}{dt}} } \]

Equivalently, by the chain rule,

\[ \frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx}. \]

The main point in this problem is the differentiation of

\[ \log\tan\frac{t}{2}. \]
We must apply the chain rule carefully.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Differentiate \(x\) with respect to \(t\).

  2. Differentiate the logarithmic term

    \[ \log\tan\frac{t}{2} \]
    using the chain rule.

  3. Differentiate \(y=\sin t\) with respect to \(t\).

  4. Apply

    \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}. \]

  5. Simplify the resulting expression.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  26 steps
  1. Differentiate \(x\) with respect to \(t\).
  2. \[x=a\left(\cos t+\log\tan\frac{t}{2}\right)\]
  3. Differentiating both sides with respect to \(t\),
    \[\frac{dx}{dt}=a\left[\frac{d}{dt}(\cos t)+\frac{d}{dt}\left(\log\tan\frac{t}{2}\right)\right]\]
  4. We know that
    \[\frac{d}{dt}(\cos t)=-\sin t\]
  5. Differentiate \(\log\tan\dfrac{t}{2}\)
  6. Let
    \[u=\tan\frac{t}{2}\]
  7. Then
    \[\frac{d}{dt}(\log u)=\frac{1}{u}\frac{du}{dt}\]
  8. Therefore,
    \[\frac{d}{dt}\left(\log\tan\frac{t}{2}\right)=\frac{1}{\tan(t/2)}\frac{d}{dt}\left(\tan\frac{t}{2}\right)\]
  9. Using the chain rule,
    \[\frac{d}{dt}\left(\tan\frac{t}{2}\right)=\sec^2\frac{t}{2}\cdot\frac{1}{2}\]
  10. Hence,
    \[\frac{d}{dt}\left(\log\tan\frac{t}{2}\right)=\frac{1}{\tan(t/2)}\cdot\frac{1}{2}\sec^2\frac{t}{2}\]
  11. Therefore,
    \[\frac{d}{dt}\left(\log\tan\frac{t}{2}\right)=\frac{1}{2}\frac{\sec^2(t/2)}{\tan(t/2)}\]
  12. Using
    \[\frac{\sec^2 u}{\tan u}=\frac{1/\cos^2u}{\sin u/\cos u}=\frac{1}{\sin u\cos u}\]
  13. with \(u=\dfrac{t}{2}\), we get
    \[\frac{d}{dt}\left(\log\tan\frac{t}{2}\right)=\frac{1}{2\sin(t/2)\cos(t/2)}\]
  14. Using the double-angle identity
    \[\sin t=2\sin\frac{t}{2}\cos\frac{t}{2}\]
  15. we obtain
    \[\boxed{\frac{d}{dt}\left(\log\tan\frac{t}{2}\right)=\frac{1}{\sin t}}\]
  16. Complete \(\dfrac{dx}{dt}\)
  17. We have
    \[\frac{dx}{dt}=a\left(-\sin t+\frac{1}{\sin t}\right)\]
  18. Taking the common denominator \(\sin t\),
    \[\frac{dx}{dt}=a\left(\frac{-\sin^2t+1}{\sin t}\right)\]
  19. Since
    \[1-\sin^2t=\cos^2t\]
  20. we get
    \[\boxed{\frac{dx}{dt}=a\frac{\cos^2t}{\sin t}}\]
  21. Differentiate \(y\) with respect to \(t\).
  22. \[y=\sin t\]
  23. Differentiating,
    \[\frac{dy}{dt}=\cos t\]
  24. Hence,
    \[\boxed{\frac{dy}{dt}=\cos t}\]
  26. \[\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\]
  27. Substituting the values,
    \[\frac{dy}{dx}=\frac{\cos t}{a\cos^2t/\sin t}\]
  28. Dividing by a fraction is equivalent to multiplying by its reciprocal:
    \[\frac{dy}{dx}=\cos t\cdot\frac{\sin t}{a\cos^2t}\]
  29. Cancel one factor of \(\cos t\):
    \[\frac{dy}{dx}=\frac{\sin t}{a\cos t}\]
  30. Therefore,
    \[\boxed{\frac{dy}{dx}=\frac{1}{a}\tan t}\]
🎯 Exam Significance
Exam Significance
›

This question is important because it combines parametric differentiation, logarithmic differentiation, the chain rule, and trigonometric identities. The most challenging part is usually the derivative of

\[ \log\tan\frac{t}{2}. \]

The key intermediate result is

\[ \frac{d}{dt} \left( \log\tan\frac{t}{2} \right) = \frac{1}{\sin t}. \]
Once this is obtained, the remaining calculation becomes straightforward.

Writing the logarithmic derivative step by step is recommended in a board examination because it demonstrates the correct use of the chain rule and makes the simplification easy to verify.

Significance for Competitive Entrance Exam Aspirants

This problem is particularly useful for competitive preparation because it teaches recognition of a non-obvious standard derivative:

\[ \frac{d}{dt} \left( \log\tan\frac{t}{2} \right) = \csc t. \]

Once this identity is recognised, the solution can be shortened considerably:

\[ \frac{dx}{dt} = a\left(-\sin t+\csc t\right) \]
\[ = a\left( \frac{1-\sin^2t}{\sin t} \right) \]
\[ = a\frac{\cos^2t}{\sin t}. \]
Since
\[ \frac{dy}{dt}=\cos t, \]
we immediately obtain
\[ \frac{dy}{dx} = \frac{\cos t} {a\cos^2t/\sin t} = \frac{\tan t}{a}. \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. For parametric equations,

    \[\boxed{\frac{dy}{dx}=\frac{dy/dt}{dx/dt}}\]

  2. The important logarithmic derivative is

    \[\boxed{\frac{d}{dt}\left(\log\tan\frac{t}{2}\right)=\frac{1}{\sin t}}\]

  3. Use the double-angle identity

    \[\sin t=2\sin\frac{t}{2}\cos\frac{t}{2}\]
    when simplifying the logarithmic derivative.

  4. The identity

    \[1-\sin^2t=\cos^2t\]
    simplifies \(\dfrac{dx}{dt}\).

  5. For the intended equation \(y=\sin t\),

    \[\frac{dy}{dt}=\cos t.\]

  6. The final derivative is

    \[\boxed{\frac{dy}{dx}=\frac{\tan t}{a}}\]

  7. The notation \(\operatorname{asin}t\) should not be used for \(\sin t\), because \(\operatorname{asin}t\) conventionally denotes the inverse sine function.
← Q7
8 / 11  ·  73%
Q9 →
Q9
NUMERIC3 marks
If \(x\) and \(y\) are connected parametrically by the given equations, without eliminating the parameter, find \(\frac{dy}{dx}\) \[ \begin{aligned} x &= a\sec\theta,\\ y &= b\tan\theta \end{aligned} \]
📘 Concept & Theory
Concept/Theory
›

When \(x\) and \(y\) are expressed in terms of a parameter \(\theta\), the derivative \(\dfrac{dy}{dx}\) can be found directly without eliminating the parameter.

The standard formula for parametric differentiation is

\[\boxed{\frac{dy}{dx}=\frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}}\]

Equivalently, using the chain rule,

\[\frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta}{dx}\]

This problem involves the standard trigonometric derivatives

\[\frac{d}{d\theta}(\sec\theta)=\sec\theta\tan\theta\]
and
\[\frac{d}{d\theta}(\tan\theta)=\sec^2\theta\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Differentiate \(x=a\sec\theta\) with respect to \(\theta\).

  2. Differentiate \(y=b\tan\theta\) with respect to \(\theta\).

  3. Use

    \[\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}\]

  4. Cancel the common factors and simplify the trigonometric expression.

  5. Express the final result in terms of \(\operatorname{cosec}\theta\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  21 steps
  1. Differentiate \(x\) with respect to \(\theta\)
  2. \[x=a\sec\theta\]
  3. Differentiating both sides with respect to \(\theta\),
    \[\frac{dx}{d\theta}=a\frac{d}{d\theta}(\sec\theta)\]
  4. Using
    \[\frac{d}{d\theta}(\sec\theta)=\sec\theta\tan\theta,\]
  5. we obtain
    \[\boxed{\frac{dx}{d\theta}=a\sec\theta\tan\theta}\]
  6. Therefore,
    \[\frac{d\theta}{dx}=\frac{1}{a\sec\theta\tan\theta}\]
  7. Hence,
    \[\boxed{\frac{d\theta}{dx}=\frac{1}{a\sec\theta\tan\theta}}\]
  8. Differentiate \(y\) with respect to \(\theta\)
  9. \[y=b\tan\theta\]
  10. Differentiating both sides,
    \[\frac{dy}{d\theta}=b\frac{d}{d\theta}(\tan\theta)\]
  11. Since
    \[\frac{d}{d\theta}(\tan\theta)=\sec^2\theta,\]
  12. we get
    \[\boxed{\frac{dy}{d\theta}=b\sec^2\theta}\]
  13. Apply the parametric differentiation formula
  14. \[\frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta}{dx}\]
  15. Substituting the values obtained above,
    \[\frac{dy}{dx}=b\sec^2\theta\cdot\frac{1}{a\sec\theta\tan\theta}\]
  16. Therefore,
    \[\frac{dy}{dx}=\frac{b\sec^2\theta}{a\sec\theta\tan\theta}\]
  17. Cancel one factor of \(\sec\theta\):
    \[\frac{dy}{dx}=\frac{b\sec\theta}{a\tan\theta}\]
  19. Since
    \[\sec\theta=\frac{1}{\cos\theta}\]
    and
    \[\tan\theta=\frac{\sin\theta}{\cos\theta},\]
  20. we have
    \[\frac{dy}{dx}=\frac{b}{a}\frac{\sec\theta}{\tan\theta}\]
    \[=\frac{b}{a}\frac{\dfrac{1}{\cos\theta}}{\dfrac{\sin\theta}{\cos\theta}}\]
  21. Dividing by a fraction,
    \[\frac{dy}{dx}=\frac{b}{a}\left(\frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta}\right)\]
  22. Cancel \(\cos\theta\):
    \[\frac{dy}{dx}=\frac{b}{a}\frac{1}{\sin\theta}\]
  23. Since
    \[\frac{1}{\sin\theta}=\operatorname{cosec}\theta,\]
  24. we obtain
    \[\boxed{\frac{dy}{dx}=\frac{b}{a}\operatorname{cosec}\theta}\]
🎯 Exam Significance
Exam Significance
›

This question is a direct application of parametric differentiation and tests the standard derivatives of \(\sec\theta\) and \(\tan\theta\). It is important to distinguish between

\[ \frac{d}{d\theta}(\sec\theta) = \sec\theta\tan\theta \]
and
\[ \frac{d}{d\theta}(\tan\theta) = \sec^2\theta. \]

Writing these derivatives correctly is essential for obtaining the correct final answer in a board examination.

The cancellation of \(\sec\theta\) followed by conversion to sine and cosine provides a clean route to the final expression:

\[ \frac{\sec\theta}{\tan\theta} = \operatorname{cosec}\theta. \]
Significance for Competitive Entrance Exam Aspirants

For competitive examinations, the key is to recognise the ratio immediately:

\[ \frac{dy}{dx} = \frac{b\sec^2\theta} {a\sec\theta\tan\theta}. \]

Cancelling one \(\sec\theta\) gives

\[ \frac{dy}{dx} = \frac{b}{a} \frac{\sec\theta}{\tan\theta}. \]

Since

\[ \frac{\sec\theta}{\tan\theta} = \operatorname{cosec}\theta, \]
the answer follows quickly:

\[ \boxed{ \frac{dy}{dx} = \frac{b}{a}\operatorname{cosec}\theta } \]

This pattern is useful in objective questions involving parametric curves, tangent slopes, and values of derivatives at specified parameter values.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. For parametric equations,

    \[\boxed{\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}}.\]

  2. The derivative of \(\sec\theta\) is

    \[\boxed{\frac{d}{d\theta}(\sec\theta)=\sec\theta\tan\theta}.\]

  3. The derivative of \(\tan\theta\) is

    \[\boxed{\frac{d}{d\theta}(\tan\theta)=\sec^2\theta}.\]

  4. For \(x=a\sec\theta\),

    \[\frac{dx}{d\theta}=a\sec\theta\tan\theta.\]

  5. For \(y=b\tan\theta\),

    \[\frac{dy}{d\theta}=b\sec^2\theta.\]

  6. The useful identity

    \[\frac{\sec\theta}{\tan\theta}=\operatorname{cosec}\theta\]
    simplifies the final result.

  7. The final derivative is

    \[\boxed{\frac{dy}{dx}=\frac{b}{a}\operatorname{cosec}\theta}\]
← Q8
9 / 11  ·  82%
Q10 →
Q10
NUMERIC3 marks
If \(x\) and \(y\) are connected parametrically by the given equations, without eliminating the parameter, find \(\frac{dy}{dx}\) \[\begin{aligned} x &= a\left(\cos\theta+\theta\sin\theta\right),\\ y &= a\left(\sin\theta-\theta\cos\theta\right) \end{aligned}\]
📘 Concept & Theory
Concept/Theory
›

When both \(x\) and \(y\) are expressed in terms of a parameter \(\theta\), the derivative \(\dfrac{dy}{dx}\) can be obtained without eliminating the parameter by using

\[\boxed{\frac{dy}{dx}=\frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}}\]

This follows from the chain rule:

\[\frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta}{dx}\]

The important feature of this problem is that both \(x\) and \(y\) contain products involving \(\theta\), namely

\[\theta\sin\theta\]
and
\[\theta\cos\theta\]
Therefore, the product rule must be used.

Recall the product rule:

\[\frac{d}{d\theta}[uv]=u\frac{dv}{d\theta}+v\frac{du}{d\theta}\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Differentiate \(x\) with respect to \(\theta\).

  2. Apply the product rule to \(\theta\sin\theta\).

  3. Differentiate \(y\) with respect to \(\theta\).

  4. Apply the product rule to \(\theta\cos\theta\).

  5. Substitute the two derivatives into

    \[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}. \]

  6. Cancel the common factors and simplify the result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  32 steps
  1. Differentiate \(x\) with respect to \(\theta\)
  2. \[x=a\left(\cos\theta+\theta\sin\theta\right)\]
  3. Expanding,
    \[x=a\cos\theta+a\theta\sin\theta\]
  4. Differentiating both sides with respect to \(\theta\),
    \[\frac{dx}{d\theta}=a\frac{d}{d\theta}(\cos\theta)+a\frac{d}{d\theta}(\theta\sin\theta)\]
  5. Since
    \[\frac{d}{d\theta}(\cos\theta)=-\sin\theta,\]
  6. we have
    \[\frac{dx}{d\theta}=-a\sin\theta+a\frac{d}{d\theta}(\theta\sin\theta)\]
  7. Apply the product rule to \(\theta\sin\theta\).
  8. Let
    \[u=\theta,\quad v=\sin\theta\]
  9. Then
    \[\frac{du}{d\theta}=1\]
    and
    \[\frac{dv}{d\theta}=\cos\theta\]
  10. Therefore, by the product rule,
    \[\frac{d}{d\theta}(\theta\sin\theta)=\theta\frac{d}{d\theta}(\sin\theta)+\sin\theta\frac{d\theta}{d\theta}\]
    \[=\theta\cos\theta+\sin\theta\]
  11. Hence,
    \[\boxed{\frac{d}{d\theta}(\theta\sin\theta)=\sin\theta+\theta\cos\theta}\]
  12. Substituting this into \(\dfrac{dx}{d\theta}\),
    \[\frac{dx}{d\theta}=-a\sin\theta+a(\sin\theta+\theta\cos\theta)\]
  13. Expanding,
    \[\frac{dx}{d\theta}=-a\sin\theta+a\sin\theta+a\theta\cos\theta\]
  14. The first two terms cancel:
    \[-a\sin\theta+a\sin\theta=0\]
  15. Therefore,
    \[\boxed{\frac{dx}{d\theta}=a\theta\cos\theta}\]
  16. Differentiate \(y\) with respect to \(\theta\)
    \[y=a\left(\sin\theta-\theta\cos\theta\right)\]
  17. Expanding,
    \[y=a\sin\theta-a\theta\cos\theta\]
  18. Differentiating,
    \[\frac{dy}{d\theta}=a\frac{d}{d\theta}(\sin\theta)-a\frac{d}{d\theta}(\theta\cos\theta)\]
  19. Since
    \[\frac{d}{d\theta}(\sin\theta)=\cos\theta,\]
  20. we get
    \[\frac{dy}{d\theta}=a\cos\theta-a\frac{d}{d\theta}(\theta\cos\theta)\]
  21. Apply the product rule to \(\theta\cos\theta\)
  22. Let
    \[u=\theta,\quad v=\cos\theta\]
  23. Then
    \[\frac{du}{d\theta}=1\]
    and
    \[\frac{dv}{d\theta}=-\sin\theta\]
  24. Therefore,
    \[\frac{d}{d\theta}(\theta\cos\theta)=\theta\frac{d}{d\theta}(\cos\theta)+\cos\theta\frac{d\theta}{d\theta}\]
    \[=\theta(-\sin\theta)+\cos\theta\]
    \[=\cos\theta-\theta\sin\theta\]
  25. Hence,
    \[\boxed{\frac{d}{d\theta}(\theta\cos\theta)=\cos\theta-\theta\sin\theta}\]
  26. Substituting into \(\dfrac{dy}{d\theta}\),
    \[\frac{dy}{d\theta}=a\cos\theta-a(\cos\theta-\theta\sin\theta)\]
  27. Expanding,
    \[\frac{dy}{d\theta}=a\cos\theta-a\cos\theta+a\theta\sin\theta\]
  28. The first two terms cancel:
    \[a\cos\theta-a\cos\theta=0\]
  29. Therefore,
    \[\boxed{\frac{dy}{d\theta}=a\theta\sin\theta}\]
  30. Apply the formula for parametric differentiation.
    \[\frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta}{dx}\]
  31. Equivalently,
    \[\frac{dy}{dx}=\frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}\]
  32. Substituting,
    \[\frac{dy}{dx}=\frac{a\theta\sin\theta}{a\theta\cos\theta}\]
  33. Assuming the denominator is non-zero, cancel the common factors \(a\) and \(\theta\):
    \[\frac{dy}{dx}=\frac{\sin\theta}{\cos\theta}\]
  34. Since
    \[\tan\theta=\frac{\sin\theta}{\cos\theta},\]
  35. we obtain
    \[\boxed{\frac{dy}{dx}=\tan\theta}\]
🎯 Exam Significance
Exam Significance
›

This problem is an important application of parametric differentiation because it combines the parametric derivative formula with the product rule. Students must differentiate

\[ \theta\sin\theta \]
and
\[ \theta\cos\theta \]
correctly.

A common mistake is to differentiate a product as though it were a single function. The correct product rule is

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

The cancellations in this problem are also important. After applying the product rule, the terms involving \(a\sin\theta\) cancel in \(\dfrac{dx}{d\theta}\), while the terms involving \(a\cos\theta\) cancel in \(\dfrac{dy}{d\theta}\). Recognising these cancellations makes the solution both accurate and easy to follow.

Significance for Competitive Entrance Exam Aspirants

For competitive examinations, this problem is a useful example of how a seemingly complicated parametric expression can simplify dramatically after differentiation.

Once the product rule is applied, the essential results are

\[ \frac{dx}{d\theta}=a\theta\cos\theta \]
and
\[ \frac{dy}{d\theta}=a\theta\sin\theta. \]

Therefore, the derivative can be obtained rapidly:

\[ \frac{dy}{dx} = \frac{a\theta\sin\theta} {a\theta\cos\theta} = \tan\theta. \]

This cancellation pattern is useful in objective questions and in problems involving tangents to parametrically represented curves.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. For parametric equations,

    \[\boxed{\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}}\]

  2. Use the product rule for terms such as

    \[\theta\sin\theta\quad\text{and}\quad\theta\cos\theta\]

  3. The derivative of \(\theta\sin\theta\) is

    \[\boxed{\sin\theta+\theta\cos\theta}\]

  4. The derivative of \(\theta\cos\theta\) is

    \[\boxed{\cos\theta-\theta\sin\theta}\]

  5. For the given \(x\),

    \[\frac{dx}{d\theta}=a\theta\cos\theta\]

  6. For the given \(y\),

    \[\frac{dy}{d\theta}=a\theta\sin\theta\]

  7. The common factors \(a\) and \(\theta\) cancel in the ratio wherever the denominator is non-zero.

  8. Do not confuse \(\sin\theta,\cos\theta\) with the inverse functions \(\sin^{-1}\theta,\cos^{-1}\theta\).

  9. The final derivative is

    \[\boxed{\frac{dy}{dx}=\tan\theta}\]
← Q9
10 / 11  ·  91%
Q11 →
Q11
NUMERIC3 marks
If \[x=\sqrt{a^{\sin^{-1}t}},\quad y=\sqrt{a^{\cos^{-1}t}},\] show that\[\frac{dy}{dx}=-\frac{y}{x}\]
📘 Concept & Theory
Concept/Theory
›

This is a parametric differentiation problem because both \(x\) and \(y\) are expressed in terms of the parameter \(t\).

The required formula is

\[\boxed{\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}}\]

Since the variable \(t\) occurs in the exponents, logarithmic differentiation is the most convenient method. We first rewrite the square roots as powers:

\[ x=a^{\frac12\sin^{-1}t} \]
and
\[ y=a^{\frac12\cos^{-1}t}. \]

We then take logarithms and differentiate with respect to \(t\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rewrite the square-root expressions as powers of \(a\).

  2. Take natural logarithms of both equations.

  3. Differentiate the logarithmic equations with respect to \(t\).

  4. Use the derivatives of \(\sin^{-1}t\) and \(\cos^{-1}t\).

  5. Obtain \(\dfrac{dx}{dt}\) and \(\dfrac{dy}{dt}\).

  6. Form

    \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
    and simplify.

  7. Show that the resulting expression is

    \[ -\frac{y}{x}. \]

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  28 steps
  1. Rewrite the given equations
    \[x=\sqrt{a^{\sin^{-1}t}}\]
  2. Since
    \[\sqrt{A}=A^{1/2}\]
  3. we get
    \[x=\left(a^{\sin^{-1}t}\right)^{1/2}\]
  4. Therefore,
    \[\boxed{x=a^{\frac12\sin^{-1}t}}\]
  5. Similarly,
    \[y=\sqrt{a^{\cos^{-1}t}}\]
  6. gives
    \[y=\left(a^{\cos^{-1}t}\right)^{1/2}\]
  7. and hence
    \[\boxed{y=a^{\frac12\cos^{-1}t}}\]
  8. Take logarithms of the equation for \(x\)
    \[x=a^{\frac12\sin^{-1}t}\]
  9. Taking natural logarithms on both sides,
    \[\log x=\log\left(a^{\frac12\sin^{-1}t}\right)\]
  10. Using
    \[\log(A^n)=n\log A\]
  11. we obtain
    \[\log x=\frac12\sin^{-1}t\log a\]
  12. Thus,
    \[\boxed{\log x=\frac{\log a}{2}\sin^{-1}t}\]
  13. Differentiate with respect to \(t\)
    \[\frac{1}{x}\frac{dx}{dt}=\frac{\log a}{2}\frac{d}{dt}(\sin^{-1}t)\]
  14. We know that
    \[\frac{d}{dt}(\sin^{-1}t)=\frac{1}{\sqrt{1-t^2}}.\]
  15. Therefore,
    \[\frac{1}{x}\frac{dx}{dt}=\frac{\log a}{2\sqrt{1-t^2}}\]
  16. Multiplying both sides by \(x\),
    \[\boxed{\frac{dx}{dt}=\frac{x\log a}{2\sqrt{1-t^2}}}\]
  17. Take logarithms of the equation for \(y\).
    \[y=a^{\frac12\cos^{-1}t}\]
  18. Taking natural logarithms,
    \[\log y=\log\left(a^{\frac12\cos^{-1}t}\right)\]
  19. Therefore,
    \[\boxed{\log y=\frac{\log a}{2}\cos^{-1}t}\]
  20. Differentiate with respect to \(t\)
    \[\frac{1}{y}\frac{dy}{dt}=\frac{\log a}{2}\frac{d}{dt}(\cos^{-1}t)\]
  21. We know that
    \[\frac{d}{dt}(\cos^{-1}t)=-\frac{1}{\sqrt{1-t^2}}\]
  22. Hence,
    \[\frac{1}{y}\frac{dy}{dt}=-\frac{\log a}{2\sqrt{1-t^2}}\]
  23. Multiplying both sides by \(y\),
    \[\boxed{\frac{dy}{dt}=-\frac{y\log a}{2\sqrt{1-t^2}}}\]
  24. Apply the parametric differentiation formula
    \[\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\]
  25. Substituting the values obtained above,
    \[\frac{dy}{dx}=\frac{-\dfrac{y\log a}{2\sqrt{1-t^2}}}{\dfrac{x\log a}{2\sqrt{1-t^2}}}\]
  26. Rewrite the division as multiplication by the reciprocal:
    \[\frac{dy}{dx}=-\frac{y\log a}{2\sqrt{1-t^2}}\cdot\frac{2\sqrt{1-t^2}}{x\log a}\]
  27. Cancel the common factors
    \[\log a,\quad 2,\quad\sqrt{1-t^2}\]
  28. Thus,
    \[\frac{dy}{dx}=-\frac{y}{x}\]
  29. Therefore,
    \[\boxed{\frac{dy}{dx}=-\frac{y}{x}}\]
🎯 Exam Significance
Exam Significance
›

This problem is important because it combines three Class 12 calculus techniques:

  1. Parametric differentiation.
  2. Logarithmic differentiation.
  3. Derivatives of inverse trigonometric functions.

The key observation is that the derivatives of \(\sin^{-1}t\) and \(\cos^{-1}t\) have equal magnitudes but opposite signs:

\[ \frac{d}{dt}(\sin^{-1}t) = \frac{1}{\sqrt{1-t^2}} \]
and
\[ \frac{d}{dt}(\cos^{-1}t) = -\frac{1}{\sqrt{1-t^2}}. \]

This opposite sign is ultimately responsible for the negative sign in

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

Significance for Competitive Entrance Exam Aspirants

For competitive examinations, the most efficient approach is to take logarithms immediately. There is no need to expand the complicated exponential expressions directly.

The essential logarithmic derivatives are

\[ \frac{1}{x}\frac{dx}{dt} = \frac{\log a}{2\sqrt{1-t^2}} \]
and
\[ \frac{1}{y}\frac{dy}{dt} = -\frac{\log a}{2\sqrt{1-t^2}}. \]

Their ratio immediately gives

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = -\frac{y}{x}. \]

This pattern is particularly useful in objective questions where the required answer is expressed in terms of \(x\) and \(y\), rather than in terms of the parameter \(t\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Rewrite square roots as powers:

    \[ \sqrt{a^u}=a^{u/2}. \]

  2. For variable exponents, logarithmic differentiation is an efficient method.

  3. The derivatives of inverse trigonometric functions are

    \[ \frac{d}{dt}(\sin^{-1}t) = \frac{1}{\sqrt{1-t^2}} \]
    and
    \[ \frac{d}{dt}(\cos^{-1}t) = -\frac{1}{\sqrt{1-t^2}}. \]

  4. For parametric equations,

    \[ \boxed{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} }. \]

  5. The common factors involving \(\log a\) and \(\sqrt{1-t^2}\) cancel when the ratio is formed.

  6. The opposite signs of the inverse-trigonometric derivatives produce the negative sign in the final result.

  7. The required result is

    \[ \boxed{ \frac{dy}{dx}=-\frac{y}{x} }. \]
← Q10
11 / 11  ·  100%
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NCERT Class 12 Maths Ex 5.6 Solutions
NCERT Class 12 Maths Ex 5.6 Solutions — Complete Notes & Solutions · academia-aeternum.com
NCERT Class 12 Mathematics Chapter 5: Continuity and Differentiability Exercise 5.6 focuses on the important concept of parametric differentiation, where the variables \(x\) and \(y\) are expressed in terms of a common parameter. These questions help students understand how to find \(\frac{dy}{dx}\) directly without eliminating the parameter. The exercise covers differentiation of trigonometric, algebraic, exponential and logarithmic parametric equations using the chain rule, product rule and…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Frequently Asked Questions

    Exercise 5.6 of NCERT Class 12 Mathematics Chapter 5 focuses on finding derivatives of parametric equations using parametric differentiation.

    If x and y are functions of a parameter t, then dy/dx = (dy/dt)/(dx/dt), provided dx/dt is not zero.

    Exercise 5.6 of NCERT Class 12 Mathematics Chapter 5 focuses on finding derivatives of parametric equations using parametric differentiation.

    If x and y are functions of a parameter t, then dy/dx = (dy/dt)/(dx/dt), provided dx/dt is not zero.

    Parametric differentiation allows us to find dy/dx directly when x and y are expressed in terms of a common parameter, without eliminating the parameter.

    Differentiate x and y separately with respect to the parameter using standard trigonometric derivative rules and then apply dy/dx = (dy/dt)/(dx/dt).

    The chain rule is used when a function contains a composite expression such as sin 2t, cos 2t, or a function of another variable.

    Yes. Logarithmic differentiation is useful for parametric expressions involving variable exponents, powers, exponential functions and logarithms.

    You should verify that dx/dt is defined and non-zero at the point under consideration so that the ratio (dy/dt)/(dx/dt) is valid.

    Yes. Parametric differentiation is an important application of differentiation and understanding its method helps students solve calculus questions accurately in board examinations.

    It strengthens differentiation techniques, chain rule, product rule, trigonometric differentiation and logarithmic differentiation, which are frequently required in competitive mathematics.

    First understand the parametric differentiation formula, then practise differentiating x and y separately, simplify the ratio carefully and revise common trigonometric and logarithmic derivatives.

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