Class 11 Limits & Derivatives Misc Exercise

Concept Used

The derivative from first principle is defined as:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

For trigonometric functions, the following standard limits are essential:

\[ \begin{aligned} \lim_{h \to 0} \frac{\sin h}{h} = 1, \\ \lim_{h \to 0} \frac{1 - \cos h}{h} = 0 \end{aligned} \]

Solution Roadmap

Step 1: Substitute \(f(x+h)\) into definition
Step 2: Simplify numerator algebraically
Step 3: Factor and cancel \(h\)
Step 4: Apply standard limits
Step 5: Evaluate limit

Solution

Linear function → constant slope → constant derivative

(i) \(f(x) = -x\)

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{-(x+h) - (-x)}{h} \\ &= \lim_{h \to 0} \frac{-x - h + x}{h} \\ &= \lim_{h \to 0} \frac{-h}{h} \\ &= -1 \end{aligned} \]

Hence, derivative is constant.

(ii) \(f(x) = (-x)^{-1} = -\dfrac{1}{x}\)

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{-\frac{1}{x+h} + \frac{1}{x}}{h} \\ &= \lim_{h \to 0} \frac{\frac{-x + (x+h)}{x(x+h)}}{h} \\ &= \lim_{h \to 0} \frac{h}{h \cdot x(x+h)} \\ &= \lim_{h \to 0} \frac{1}{x(x+h)} \\ &= \frac{1}{x^2} \end{aligned} \]

Careful algebraic manipulation is critical here.

(iii) \(f(x) = \sin(x+1)\)

Shifted sine curve → derivative remains cosine

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\sin(x+1+h) - \sin(x+1)}{h} \\ \end{aligned} \]

Using identity:

\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \]

\[ \begin{aligned} &= \lim_{h \to 0} \frac{2 \cos\left(x+1+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} \\ &= \lim_{h \to 0} \cos\left(x+1+\frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h/2} \\ &= \cos(x+1) \end{aligned} \]

(iv) \(f(x) = \cos\left(x - \frac{\pi}{8}\right)\)

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\cos\left(x - \frac{\pi}{8} + h\right) - \cos\left(x - \frac{\pi}{8}\right)}{h} \end{aligned} \]

Using identity:

\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \]

\[ \begin{aligned} &= \lim_{h \to 0} \frac{-2 \sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} \\ &= \lim_{h \to 0} -\sin\left(x - \frac{\pi}{8} + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h/2} \\ &= -\sin\left(x - \frac{\pi}{8}\right) \end{aligned} \]

Exam Significance

Directly tests understanding of first principle (very common in CBSE boards)
Frequently asked in JEE/NEET as conceptual MCQs
Forms base for all derivative rules (chain rule, product rule)
Trigonometric limits are extremely important for competitive exams

Concept Used

The derivative measures the rate of change of a function with respect to a variable. Using first principle:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

Important facts:

Derivative of variable \(x\) is \(1\)
Derivative of constant is \(0\)
Derivative of sum = sum of derivatives

Solution Roadmap

Step 1: Apply first principle definition
Step 2: Substitute \(f(x+h)\)
Step 3: Simplify numerator
Step 4: Cancel \(h\)
Step 5: Take limit

Solution

Straight line with constant slope = 1 (independent of \(a\))

Let the function be:

\[ f(x) = x + a \]

Using first principle:

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \end{aligned} \]

Now compute \(f(x+h)\):

\[ f(x+h) = (x+h) + a \]

Substitute into definition:

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{(x+h+a) - (x+a)}{h} \end{aligned} \]

Simplify numerator:

\[ \begin{aligned} &= \lim_{h \to 0} \frac{x + h + a - x - a}{h} \\ &= \lim_{h \to 0} \frac{h}{h} \end{aligned} \]

Cancel \(h\):

\[ = \lim_{h \to 0} 1 = 1 \]

Hence,

\[ f'(x) = 1 \]

Key Insight

The constant \(a\) only shifts the graph vertically and does not affect slope. Therefore, derivative depends only on \(x\), not on \(a\).

Exam Significance

Fundamental concept used in almost every derivative problem
Frequently appears in assertion-reason and MCQs in JEE/NEET
Tests understanding of constants vs variables
Base case for linear functions \(ax+b\)

Concept Used

This problem combines algebraic simplification with differentiation.

Expand product before differentiating (reduces complexity)
Derivative of constant = 0
\(\dfrac{d}{dx}(x) = 1\)
\(\dfrac{d}{dx}\left(x^{-1}\right) = -x^{-2}\)

This avoids direct use of product rule and simplifies computation.

Solution Roadmap

Step 1: Expand the expression completely
Step 2: Rearrange into standard polynomial + power terms
Step 3: Differentiate term-wise
Step 4: Simplify final expression

Solution

Combination of linear and reciprocal terms → mixed derivative behavior

Let the function be:

\[ f(x) = (px+q)\left(\dfrac{r}{x}+s\right) \]

First, expand the expression:

\[ \begin{aligned} f(x) &= px \cdot \frac{r}{x} + px \cdot s + q \cdot \frac{r}{x} + q \cdot s \end{aligned} \]

Simplify each term carefully:

\[ \begin{aligned} px \cdot \frac{r}{x} &= pr \quad (\text{since } x \text{ cancels}) \\ px \cdot s &= psx \\ q \cdot \frac{r}{x} &= \frac{qr}{x} \\ q \cdot s &= qs \end{aligned} \]

Thus,

\[ \begin{aligned} f(x) &= pr + psx + \frac{qr}{x} + qs \end{aligned} \]

Rearrange:

\[ f(x) = (pr + qs) + psx + \frac{qr}{x} \]

Now differentiate term-wise:

\[ \begin{aligned} f'(x) &= \frac{d}{dx}(pr + qs) + \frac{d}{dx}(psx) + \frac{d}{dx}\left(\frac{qr}{x}\right) \end{aligned} \]

Evaluate each derivative:

\[ \begin{aligned} \frac{d}{dx}(pr + qs) &= 0 \quad (\text{constant}) \\ \frac{d}{dx}(psx) &= ps \\ \frac{d}{dx}\left(\frac{qr}{x}\right) &= qr \cdot \frac{d}{dx}(x^{-1}) = -\frac{qr}{x^2} \end{aligned} \]

Hence,

\[ \begin{aligned} f'(x) &= 0 + ps - \frac{qr}{x^2} \\ &= ps - \frac{qr}{x^2} \end{aligned} \]

Key Insight

Expanding before differentiation avoids product rule and reduces chances of error. This is a standard strategy in algebra-heavy derivative problems.

Exam Significance

Tests algebraic manipulation + differentiation together
Very common in CBSE board subjective questions
In JEE/NEET, appears as simplification-based MCQ
Builds foundation for product rule understanding

Concept Used

Algebraic expansion of polynomial expressions
Power rule: \(\dfrac{d}{dx}(x^n) = nx^{n-1}\)
Derivative of constant = 0

Instead of using product rule, expanding simplifies differentiation into basic polynomial form.

Solution Roadmap

Step 1: Expand \((cx+d)^2\)
Step 2: Multiply with \((ax+b)\)
Step 3: Combine like terms
Step 4: Differentiate term-wise

Solution

Product of linear and quadratic → cubic function → quadratic derivative

Let the function be:

\[ f(x) = (ax+b)(cx+d)^2 \]

Step 1: Expand \((cx+d)^2\)

\[ (cx+d)^2 = c^2x^2 + 2cdx + d^2 \]

Step 2: Multiply with \((ax+b)\)

\[ \begin{aligned} f(x) &= (ax+b)(c^2x^2 + 2cdx + d^2) \end{aligned} \]

Distribute carefully:

\[ \begin{aligned} f(x) &= ax \cdot c^2x^2 + ax \cdot 2cdx + ax \cdot d^2 \\ &\quad + b \cdot c^2x^2 + b \cdot 2cdx + b \cdot d^2 \end{aligned} \]

Simplify each term:

\[ \begin{aligned} ax \cdot c^2x^2 &= ac^2x^3 \\ ax \cdot 2cdx &= 2acdx^2 \\ ax \cdot d^2 &= ad^2x \\ b \cdot c^2x^2 &= bc^2x^2 \\ b \cdot 2cdx &= 2bcdx \\ b \cdot d^2 &= bd^2 \end{aligned} \]

Step 3: Combine like terms

\[ \begin{aligned} f(x) &= ac^2x^3 + (2acd + bc^2)x^2 + (ad^2 + 2bcd)x + bd^2 \end{aligned} \]

Step 4: Differentiate term-wise

\[ \begin{aligned} f'(x) &= \frac{d}{dx}(ac^2x^3) + \frac{d}{dx}((2acd + bc^2)x^2) \\ &\quad + \frac{d}{dx}((ad^2 + 2bcd)x) + \frac{d}{dx}(bd^2) \end{aligned} \]

Now compute each derivative:

\[ \begin{aligned} \frac{d}{dx}(ac^2x^3) &= 3ac^2x^2 \\ \frac{d}{dx}((2acd + bc^2)x^2) &= 2(2acd + bc^2)x \\ \frac{d}{dx}((ad^2 + 2bcd)x) &= (ad^2 + 2bcd) \\ \frac{d}{dx}(bd^2) &= 0 \end{aligned} \]

Final Answer:

\[ f'(x) = 3ac^2x^2 + 2(2acd + bc^2)x + (ad^2 + 2bcd) \]

Key Insight

Expanding converts a complex product into a simple polynomial. This avoids product rule and reduces calculation errors in exam settings.

Exam Significance

Classic CBSE long-answer question type
Tests algebra + differentiation simultaneously
In JEE/NEET, used to check simplification strategy
Foundation for understanding product rule efficiency

Concept Used

Quotient Rule of Differentiation
If \( f(x) = \dfrac{u(x)}{v(x)} \), then

\[ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \]

Derivative of linear function \(ax+b\) is \(a\)
Derivative of \(cx+d\) is \(c\)

Solution Roadmap

Step 1: Identify numerator \(u(x)\) and denominator \(v(x)\)
Step 2: Compute \(u'(x)\) and \(v'(x)\)
Step 3: Apply quotient rule carefully
Step 4: Expand and simplify numerator
Step 5: Write final simplified form

Solution

Rational function → slope depends on denominator squared

Let

\[ u(x) = ax + b, \quad v(x) = cx + d \]

Then,

\[ u'(x) = a, \quad v'(x) = c \]

Using quotient rule:

\[ \begin{aligned} f'(x) &= \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \end{aligned} \]

Substitute values:

\[ \begin{aligned} f'(x) &= \frac{(cx+d)(a) - (ax+b)(c)}{(cx+d)^2} \end{aligned} \]

Now expand numerator carefully:

\[ \begin{aligned} (cx+d)(a) &= acx + ad \\ (ax+b)(c) &= acx + bc \end{aligned} \]

Substitute back:

\[ \begin{aligned} f'(x) &= \frac{acx + ad - (acx + bc)}{(cx+d)^2} \end{aligned} \]

Remove brackets:

\[ \begin{aligned} &= \frac{acx + ad - acx - bc}{(cx+d)^2} \end{aligned} \]

Simplify:

\[ \begin{aligned} &= \frac{ad - bc}{(cx+d)^2} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{ad - bc}{(cx+d)^2} \]

Key Insight

Notice that the \(x\)-terms cancel completely in the numerator. This is a very important pattern: derivative becomes a constant divided by \((cx+d)^2\).

Common Mistake

Students often forget the minus sign in quotient rule or fail to expand properly, leading to incorrect cancellation.

Exam Significance

Very common identity-type question in CBSE boards
Frequently used in JEE MCQs (direct formula recognition)
Forms base for inverse trigonometric derivatives later
Important for understanding rational function behavior

Concept Used

Algebraic simplification before differentiation
Quotient Rule
Derivative of linear functions

Rational expressions involving \(\frac{1}{x}\) are often simplified by multiplying numerator and denominator by \(x\).

Solution Roadmap

Step 1: Eliminate fractions by multiplying by \(x\)
Step 2: Convert into simple rational form
Step 3: Apply quotient rule
Step 4: Expand and simplify carefully

Solution

Vertical asymptote at \(x=1\) → derivative depends on \((x-1)^2\)

Step 1: Simplify the function

\[ \begin{aligned} f(x) &= \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} \end{aligned} \]

Multiply numerator and denominator by \(x\):

\[ \begin{aligned} f(x) &= \frac{x\left(1 + \frac{1}{x}\right)}{x\left(1 - \frac{1}{x}\right)} \\ &= \frac{x + 1}{x - 1} \end{aligned} \]

Step 2: Apply quotient rule

Let \(u = x+1\), \(v = x-1\)

\[ u' = 1, \quad v' = 1 \]

\[ f'(x) = \frac{v u' - u v'}{v^2} \]

Substitute values:

\[ \begin{aligned} f'(x) &= \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2} \end{aligned} \]

Step 3: Simplify numerator

\[ \begin{aligned} &= \frac{x - 1 - x - 1}{(x-1)^2} \\ &= \frac{-2}{(x-1)^2} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{-2}{(x-1)^2} \]

Key Insight

Simplification before differentiation reduces complexity drastically. Without simplifying, quotient rule would involve fractions and become error-prone.

Common Mistake

Students often directly apply quotient rule to the original form, leading to messy algebra and higher chances of sign errors.

Exam Significance

Classic simplification-based derivative problem
Frequently appears in CBSE board exams
Common in JEE as a trap question (tests simplification skill)
Builds habit of reducing expressions before differentiation

Concept Used

Chain Rule (most efficient method)
Rewrite reciprocal as power: \((ax^2+bx+c)^{-1}\)
Derivative of \(x^n = nx^{n-1}\)

Although quotient rule works, chain rule is faster and conceptually cleaner here.

Solution Roadmap

Step 1: Rewrite function in power form
Step 2: Apply chain rule
Step 3: Differentiate inner function
Step 4: Combine results carefully

Solution

Reciprocal of quadratic → slope involves derivative of denominator

Step 1: Rewrite function

\[ f(x) = (ax^2 + bx + c)^{-1} \]

Step 2: Apply chain rule

\[ \frac{d}{dx}(u^{-1}) = -u^{-2} \cdot \frac{du}{dx} \]

Let \(u = ax^2 + bx + c\)

\[ \frac{du}{dx} = 2ax + b \]

Step 3: Substitute

\[ \begin{aligned} f'(x) &= -(ax^2 + bx + c)^{-2} \cdot (2ax + b) \end{aligned} \]

Step 4: Write in fraction form

\[ f'(x) = \frac{-(2ax + b)}{(ax^2 + bx + c)^2} \]

Final Answer:

\[ f'(x) = \frac{-(2ax + b)}{(ax^2 + bx + c)^2} \]

Alternative Method (Quotient Rule)

Taking \(f(x) = \dfrac{1}{ax^2+bx+c}\), apply quotient rule:

\[ f'(x) = \frac{(ax^2+bx+c)(0) - 1(2ax+b)}{(ax^2+bx+c)^2} \]

This also gives the same result.

Key Insight

Whenever you see \(\frac{1}{f(x)}\), think of rewriting it as \(f(x)^{-1}\). This immediately suggests chain rule and simplifies the process.

Common Mistake

Missing the negative sign or forgetting to differentiate the inner function \((2ax+b)\).

Exam Significance

Very common in CBSE board exams
Frequently appears in JEE as a chain rule application
Important for rational and inverse function derivatives
Builds strong foundation for advanced calculus

Concept Used

Quotient Rule
Derivative of linear and quadratic polynomials
Careful algebraic expansion and simplification

For \( f(x)=\dfrac{u(x)}{v(x)} \), \[ f'(x)=\frac{v u' - u v'}{v^2} \]

Solution Roadmap

Step 1: Identify \(u(x)\) and \(v(x)\)
Step 2: Compute derivatives \(u'(x)\), \(v'(x)\)
Step 3: Apply quotient rule
Step 4: Expand numerator carefully
Step 5: Combine like terms

Solution

Linear over quadratic → derivative becomes rational function with squared denominator

Step 1: Define functions

\[ u(x) = ax + b, \quad v(x) = px^2 + qx + r \]

\[ u'(x) = a, \quad v'(x) = 2px + q \]

Step 2: Apply quotient rule

\[ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \]

Substitute values:

\[ \begin{aligned} f'(x) &= \frac{(px^2 + qx + r)(a) - (ax + b)(2px + q)}{(px^2 + qx + r)^2} \end{aligned} \]

Step 3: Expand each part

\[ \begin{aligned} (px^2 + qx + r)(a) &= apx^2 + aqx + ar \end{aligned} \]

\[ \begin{aligned} (ax + b)(2px + q) &= ax(2px + q) + b(2px + q) \\ &= 2apx^2 + aqx + 2bpx + bq \end{aligned} \]

Step 4: Substitute back

\[ \begin{aligned} f'(x) &= \frac{apx^2 + aqx + ar - (2apx^2 + aqx + 2bpx + bq)}{(px^2 + qx + r)^2} \end{aligned} \]

Step 5: Remove brackets

\[ \begin{aligned} &= \frac{apx^2 + aqx + ar - 2apx^2 - aqx - 2bpx - bq}{(px^2 + qx + r)^2} \end{aligned} \]

Step 6: Combine like terms

\[ \begin{aligned} &= \frac{-apx^2 - 2bpx + ar - bq}{(px^2 + qx + r)^2} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{-apx^2 - 2bpx + ar - bq}{(px^2 + qx + r)^2} \]

Key Insight

Notice that the middle term \(aqx\) cancels completely. Such cancellations are common in quotient rule and help simplify the final result.

Common Mistake

Forgetting to distribute the negative sign across all terms in \((ax+b)(2px+q)\) is a frequent error.

Exam Significance

Standard CBSE long-answer question type
Appears in JEE as algebra-heavy MCQ
Tests precision in expansion and sign handling
Foundation for advanced rational function derivatives

Concept Used

Quotient Rule
Derivative of quadratic and linear polynomials
Accurate algebraic expansion and simplification

If \( f(x)=\dfrac{u(x)}{v(x)} \), then \[ f'(x)=\frac{v u' - u v'}{v^2} \]

Solution Roadmap

Step 1: Identify \(u(x)\), \(v(x)\)
Step 2: Compute \(u'(x)\), \(v'(x)\)
Step 3: Apply quotient rule
Step 4: Expand numerator carefully
Step 5: Combine like terms

Solution

Quadratic over linear → rational curve with vertical asymptote

Step 1: Define functions

\[ \begin{aligned} u(x) &= px^2 + qx + r, \\ v(x) &= ax + b \end{aligned} \]

\[ \begin{aligned} u'(x) &= 2px + q, \\ v'(x) &= a \end{aligned} \]

Step 2: Apply quotient rule

\[ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \]

Substitute values:

\[ \begin{aligned} f'(x) &= \frac{(ax+b)(2px+q) - (px^2+qx+r)(a)}{(ax+b)^2} \end{aligned} \]

Step 3: Expand each part carefully

\[ \begin{aligned} (ax+b)(2px+q) &= ax(2px+q) + b(2px+q) \\ &= 2apx^2 + aqx + 2bpx + bq \end{aligned} \]

\[ \begin{aligned} (px^2+qx+r)(a) &= apx^2 + aqx + ar \end{aligned} \]

Step 4: Substitute back

\[ \begin{aligned} f'(x) &= \frac{2apx^2 + aqx + 2bpx + bq - (apx^2 + aqx + ar)}{(ax+b)^2} \end{aligned} \]

Step 5: Remove brackets

\[ \begin{aligned} &= \frac{2apx^2 + aqx + 2bpx + bq - apx^2 - aqx - ar}{(ax+b)^2} \end{aligned} \]

Step 6: Combine like terms

\[ \begin{aligned} &= \frac{apx^2 + 2bpx + bq - ar}{(ax+b)^2} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{apx^2 + 2bpx + bq - ar}{(ax+b)^2} \]

Key Insight

The \(aqx\) terms cancel out completely. Spotting such cancellations early can help verify correctness during exams.

Common Mistake

Missing the negative sign while subtracting \((px^2+qx+r)a\) leads to incorrect results.

Exam Significance

Mirror form of Q8 — tests conceptual symmetry
Common CBSE board long-answer problem
Appears in JEE as algebra-heavy MCQ
Strengthens quotient rule mastery

Concept Used

Power Rule: \(\dfrac{d}{dx}(x^n)=nx^{n-1}\)
Derivative of trigonometric function: \(\dfrac{d}{dx}(\cos x)=-\sin x\)
Derivative of sum = sum of derivatives

Convert fractions into power form before differentiating.

Solution Roadmap

Step 1: Rewrite terms in power form
Step 2: Differentiate each term separately
Step 3: Apply power rule and trigonometric derivative
Step 4: Rewrite final answer in standard form

Solution

Combination of algebraic decay terms and trigonometric oscillation

Step 1: Rewrite in power form

\[ f(x) = a x^{-4} - b x^{-2} + \cos x \]

Step 2: Differentiate term-wise

\[ \begin{aligned} f'(x) &= \frac{d}{dx}(a x^{-4}) - \frac{d}{dx}(b x^{-2}) + \frac{d}{dx}(\cos x) \end{aligned} \]

Step 3: Apply power rule

\[ \begin{aligned} \frac{d}{dx}(a x^{-4}) &= a(-4)x^{-5} \\&= -4ax^{-5} \\ \frac{d}{dx}(b x^{-2}) &= b(-2)x^{-3} \\&= -2bx^{-3} \end{aligned} \]

Since there is a minus sign before the second term:

\[ -\frac{d}{dx}(b x^{-2}) = -(-2bx^{-3}) = 2bx^{-3} \]

Also,

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

Step 4: Combine all terms

\[ \begin{aligned} f'(x) &= -4ax^{-5} + 2bx^{-3} - \sin x \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{-4a}{x^5} + \frac{2b}{x^3} - \sin x \]

Key Insight

Always convert fractions into powers before differentiating. This transforms complex expressions into simple applications of power rule.

Common Mistake

Missing the negative sign in \(\dfrac{d}{dx}(\cos x)=-\sin x\) or mishandling the minus before \(\dfrac{b}{x^2}\).

Exam Significance

Basic but very important derivative question
Frequently appears in CBSE boards (short answer)
Common in JEE/NEET for speed testing
Tests clarity in power rule and trig derivatives

Concept Used

Power Rule: \(\dfrac{d}{dx}(x^n)=nx^{n-1}\)
Derivative of constant = 0
Rewrite radical form into exponent form

\[ \sqrt{x} = x^{1/2} \]

Solution Roadmap

Step 1: Convert square root into power form
Step 2: Apply power rule
Step 3: Simplify exponent
Step 4: Rewrite in radical form

Solution

Square root function → decreasing slope as \(x\) increases

Step 1: Rewrite the function

\[ f(x) = 4\sqrt{x} - 2 = 4x^{1/2} - 2 \]

Step 2: Differentiate term-wise

\[ \begin{aligned} f'(x) &= \frac{d}{dx}(4x^{1/2}) - \frac{d}{dx}(2) \end{aligned} \]

Step 3: Apply power rule

\[ \begin{aligned} \frac{d}{dx}(4x^{1/2}) &= 4 \cdot \frac{1}{2} x^{1/2 - 1} \\ &= 2x^{-1/2} \end{aligned} \]

Also,

\[ \frac{d}{dx}(2) = 0 \]

Step 4: Combine

\[ f'(x) = 2x^{-1/2} \]

Step 5: Convert back to radical form

\[ f'(x) = \frac{2}{\sqrt{x}} \]

Final Answer:

\[ f'(x) = \frac{2}{\sqrt{x}} \]

Key Insight

Converting radicals to powers simplifies differentiation and avoids mistakes.

Common Mistake

Students often forget that exponent decreases by 1, leading to incorrect power \(x^{1/2}\) instead of \(x^{-1/2}\).

Exam Significance

Very common CBSE board short question
Frequently appears in JEE/NEET speed-based sections
Tests understanding of fractional powers
Foundation for differentiation of radicals

Concept Used

Chain Rule: derivative of composite functions
Power Rule: \(\dfrac{d}{dx}(x^n)=nx^{n-1}\)

If \(y = [g(x)]^n\), then \[ \frac{dy}{dx} = n[g(x)]^{n-1} \cdot g'(x) \]

Solution Roadmap

Step 1: Identify inner function \(g(x) = ax+b\)
Step 2: Differentiate outer function
Step 3: Multiply by derivative of inner function
Step 4: Simplify result

Solution

Composite function: outer power acting on linear inner function

Step 1: Define the function

\[ f(x) = (ax+b)^n \]

Step 2: Identify inner function

\[ g(x) = ax + b \]

\[ \frac{d}{dx}(g(x)) = a \]

Step 3: Apply chain rule

\[ \begin{aligned} f'(x) &= \frac{d}{dx}(g(x))^n \\ &= n(g(x))^{n-1} \cdot g'(x) \end{aligned} \]

Step 4: Substitute values

\[ \begin{aligned} f'(x) &= n(ax+b)^{n-1} \cdot a \\ &= an(ax+b)^{n-1} \end{aligned} \]

Final Answer:

\[ f'(x) = an(ax+b)^{n-1} \]

Key Insight

Always identify inner and outer functions. This is the foundation of chain rule, which is one of the most important tools in calculus.

Quick Recognition Trick

For \((ax+b)^n\), directly write:
Multiply by \(n\), reduce power by 1, and multiply by derivative of inside (\(a\)).

Common Mistake

Forgetting to multiply by derivative of inner function (\(a\)) — this leads to incomplete answers.

Exam Significance

Extremely important for CBSE board exams
Core concept for JEE/NEET calculus
Foundation for all composite functions
Frequently used in advanced differentiation problems

Concept Used

Product Rule: \(\dfrac{d}{dx}(uv)=u'v+uv'\)
Chain Rule for composite functions

Both functions are composite, so product rule + chain rule are used together.

Solution Roadmap

Step 1: Identify \(u(x)\) and \(v(x)\)
Step 2: Differentiate each using chain rule
Step 3: Apply product rule
Step 4: Factor common terms

Solution

Product of two composite functions → derivative combines both rates of change

Step 1: Define functions

\[ \begin{aligned} u(x) = (ax+b)^n, \\ v(x) = (cx+d)^m \end{aligned} \]

Step 2: Differentiate each using chain rule

\[ \begin{aligned} u'(x) &= n(ax+b)^{n-1} \cdot a \\&= na(ax+b)^{n-1} \\ v'(x) &= m(cx+d)^{m-1} \cdot c \\&= mc(cx+d)^{m-1} \end{aligned} \]

Step 3: Apply product rule

\[ \begin{aligned} f'(x) &= u'(x)v(x) + u(x)v'(x) \\ &= \left[na(ax+b)^{n-1}\right](cx+d)^m \\ &\quad + (ax+b)^n\left[mc(cx+d)^{m-1}\right] \end{aligned} \]

Step 4: Write clearly

\[ \begin{aligned} f'(x) &= na(ax+b)^{n-1}(cx+d)^m \\ &\quad + mc(ax+b)^n(cx+d)^{m-1} \end{aligned} \]

Step 5: Factor common terms

Common factor:

\[ (ax+b)^{n-1}(cx+d)^{m-1} \]

Factor it out:

\[ \begin{aligned} f'(x) &= (ax+b)^{n-1}(cx+d)^{m-1} \Big[ na(cx+d) + mc(ax+b) \Big] \end{aligned} \]

Final Answer:

\[ f'(x) = (ax+b)^{n-1}(cx+d)^{m-1} \left[ na(cx+d) + mc(ax+b) \right] \]

Key Insight

After applying product rule, factorization simplifies the result significantly. This is a standard exam technique to present answers neatly.

Quick Recognition Trick

Reduce powers by 1 for both terms and multiply:
first term by \(na(cx+d)\), second by \(mc(ax+b)\).

Common Mistake

Missing chain rule factors \(a\) and \(c\), or forgetting to reduce powers correctly.

Exam Significance

High-weight conceptual question in CBSE boards
Very important for JEE/NEET (product + chain rule combo)
Frequently used in advanced calculus problems
Builds multi-rule differentiation fluency

Concept Used

Chain Rule
Derivative of trigonometric function: \(\dfrac{d}{dx}(\sin x)=\cos x\)

If \(y = \sin(g(x))\), then \[ \frac{dy}{dx} = \cos(g(x)) \cdot g'(x) \]

Solution Roadmap

Step 1: Identify inner function \(g(x)=x+a\)
Step 2: Differentiate outer function
Step 3: Multiply by derivative of inner function
Step 4: Simplify result

Solution

Horizontal shift by \(a\) → shape unchanged → derivative unchanged in form

Step 1: Define the function

\[ f(x) = \sin(x+a) \]

Step 2: Identify inner function

\[ g(x) = x + a \]

\[ \frac{d}{dx}(g(x)) = \frac{d}{dx}(x+a) = 1 \]

Step 3: Apply chain rule

\[ \begin{aligned} f'(x) &= \frac{d}{dx}\sin(g(x)) \\ &= \cos(g(x)) \cdot g'(x) \end{aligned} \]

Step 4: Substitute values

\[ \begin{aligned} f'(x) &= \cos(x+a) \cdot 1 \\ &= \cos(x+a) \end{aligned} \]

Final Answer:

\[ f'(x) = \cos(x+a) \]

Key Insight

Adding a constant inside a function shifts the graph horizontally but does not change the derivative structure.

Quick Recognition Trick

For \(\sin(x+a)\), directly write derivative as \(\cos(x+a)\) since derivative of inner term is 1.

Common Mistake

Some students incorrectly write \(\cos x\) instead of \(\cos(x+a)\), forgetting to retain the full argument.

Exam Significance

Very common CBSE board question
Frequently appears in JEE/NEET MCQs
Tests understanding of shifted functions
Foundation for trig composite differentiation

Concept Used

Product Rule: \(\dfrac{d}{dx}(uv)=u'v+uv'\)
Trigonometric derivatives:

\[ \begin{aligned} \frac{d}{dx}(\text{cosec}\ x) &= -\text{cosec}\ x \cot x,\\ \frac{d}{dx}(\cot x) &= -\text{cosec}^2 x \end{aligned} \]

Solution Roadmap

Step 1: Identify \(u=\text{cosec}\ x\), \(v=\cot x\)
Step 2: Differentiate each function
Step 3: Apply product rule
Step 4: Simplify expression

Solution

Product of cosec and cot → rapid variation near asymptotes

Step 1: Define functions

\[ u = \text{cosec}\ x, \quad v = \cot x \]

Step 2: Differentiate each

\[ \begin{aligned} u' &= -\text{cosec}\ x \cot x \\ v' &= -\text{cosec}^2\ x \end{aligned} \]

Step 3: Apply product rule

\[ \begin{aligned} f'(x) &= u'v + uv' \\ &= (-\text{cosec}\ x \cot x)(\cot x) + (\text{cosec}\ x)(-\text{cosec}^2\ x) \end{aligned} \]

Step 4: Simplify

\[ \begin{aligned} f'(x) &= -\text{cosec}\ x \cot^2 x - \text{cosec}^3\ x \end{aligned} \]

Final Answer:

\[ f'(x) = -\text{cosec}\ x \cot^2 x - \text{cosec}^3\ x \]

Optional Simplification Insight

Using identity:

\[ \text{cosec}^2\ x = 1 + \cot^2 x \]

Expression can also be written as:

\[ f'(x) = -\text{cosec}\ x(\cot^2 x + \text{cosec}^2\ x) \]

Key Insight

Trigonometric derivatives often produce higher powers (like \(\text{cosec}^3 x\)). Recognizing identities helps simplify results further.

Common Mistake

Forgetting negative signs in derivatives of \(\text{cosec} x\) and \(\cot x\), or mixing up trig derivative formulas.

Exam Significance

Important CBSE board problem (product rule + trig)
Common in JEE/NEET conceptual MCQs
Tests trig derivative memory + algebraic handling
Foundation for complex trigonometric differentiation

Concept Used

Quotient Rule
Trigonometric derivatives
Fundamental identity: \(\sin^2 x + \cos^2 x = 1\)

If \( f(x)=\dfrac{u(x)}{v(x)} \), then \[ f'(x)=\frac{v u' - u v'}{v^2} \]

Solution Roadmap

Step 1: Identify \(u=\cos x\), \(v=a+\sin x\)
Step 2: Compute derivatives
Step 3: Apply quotient rule
Step 4: Expand numerator carefully
Step 5: Use identity to simplify

Solution

Ratio of trig functions → simplification often uses identities

Step 1: Define functions

\[ u = \cos x, \quad v = a + \sin x \]

Step 2: Differentiate

\[ u' = -\sin x, \quad v' = \cos x \]

Step 3: Apply quotient rule

\[ f'(x) = \frac{v u' - u v'}{v^2} \]

Substitute values:

\[ \begin{aligned} f'(x) &= \frac{(a+\sin x)(-\sin x) - (\cos x)(\cos x)}{(a+\sin x)^2} \end{aligned} \]

Step 4: Expand numerator

\[ \begin{aligned} &= \frac{-a\sin x - \sin^2 x - \cos^2 x}{(a+\sin x)^2} \end{aligned} \]

Step 5: Use identity

\[ \sin^2 x + \cos^2 x = 1 \]

\[ \begin{aligned} f'(x) &= \frac{-a\sin x - 1}{(a+\sin x)^2} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{-a\sin x - 1}{(a+\sin x)^2} \]

Key Insight

The identity \(\sin^2 x + \cos^2 x = 1\) is crucial here. Without it, the expression remains unnecessarily complicated.

Quick Recognition Trick

Whenever both \(\sin^2 x\) and \(\cos^2 x\) appear together, immediately replace them with 1.

Common Mistake

Forgetting to square the denominator or missing the identity step leads to incomplete simplification.

Exam Significance

Very common CBSE board problem
Frequently appears in JEE/NEET as identity-based simplification
Tests quotient rule + trig identity simultaneously
Important for advanced trigonometric differentiation

Concept Used

Quotient Rule
Trigonometric derivatives
Identities: \(\sin^2 x + \cos^2 x = 1\)
Algebraic identities: \((a \pm b)^2\)

\[ \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v u' - u v'}{v^2} \]

Solution Roadmap

Step 1: Define numerator and denominator
Step 2: Differentiate each part
Step 3: Apply quotient rule
Step 4: Convert expressions into squares
Step 5: Use identities to simplify

Solution

Symmetric trig ratio → strong cancellation patterns in derivative

Step 1: Define functions

\[ u = \sin x + \cos x,\quad v = \sin x - \cos x \]

Step 2: Differentiate

\[ \begin{aligned} u' &= \cos x - \sin x \\ v' &= \cos x + \sin x \end{aligned} \]

Step 3: Apply quotient rule

\[ f'(x) = \frac{v u' - u v'}{v^2} \]

Substitute values:

\[ \begin{aligned} f'(x) &= \frac{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2} \end{aligned} \]

Step 4: Convert into squares

\[ \begin{aligned} (\sin x - \cos x)(\cos x - \sin x) &= -(\sin x - \cos x)^2 \\ (\sin x + \cos x)(\cos x + \sin x) &= (\sin x + \cos x)^2 \end{aligned} \]

Thus, numerator becomes:

\[ \begin{aligned} &= -(\sin x - \cos x)^2 - (\sin x + \cos x)^2 \end{aligned} \]

Step 5: Expand both squares

\[ \begin{aligned} (\sin x - \cos x)^2 &= \sin^2 x + \cos^2 x - 2\sin x \cos x \ (\sin x + \cos x)^2 &= \sin^2 x + \cos^2 x + 2\sin x \cos x \end{aligned} \]

Add them:

\[ \begin{aligned} &= 2(\sin^2 x + \cos^2 x) \end{aligned} \]

Using identity:

\[ \sin^2 x + \cos^2 x = 1 \]

So numerator becomes:

\[ -2 \]

Final Answer:

\[ f'(x) = \frac{-2}{(\sin x - \cos x)^2} \]

Key Insight

Symmetric expressions like \((\sin x \pm \cos x)\) often reduce using square identities. Recognizing this avoids lengthy expansion errors.

Quick Recognition Trick

When you see \((\sin x+\cos x)/(\sin x-\cos x)\), expect cancellation leading to a constant numerator after differentiation.

Common Mistake

Not recognizing \((\cos x - \sin x) = -(\sin x - \cos x)\), which simplifies the first product.

Exam Significance

Very important CBSE board derivation-type question
Common JEE MCQ (identity + quotient rule)
Tests algebraic pattern recognition
Improves speed in trig simplification

Concept Used

Quotient Rule
Trigonometric derivatives:

\[ \frac{d}{dx}(\sec x) = \sec x \tan x \]

If \( f(x)=\dfrac{u}{v} \), then \[ f'(x)=\frac{v u' - u v'}{v^2} \]

Solution Roadmap

Step 1: Identify numerator and denominator
Step 2: Differentiate each
Step 3: Apply quotient rule
Step 4: Factor common term
Step 5: Simplify expression

Solution

Step 1: Define functions

\[ u = \sec x - 1,\quad v = \sec x + 1 \]

Step 2: Differentiate

\[ \begin{aligned} u' = \sec x \tan x,\\ v' = \sec x \tan x \end{aligned} \]

Step 3: Apply quotient rule

\[ f'(x) = \frac{v u' - u v'}{v^2} \]

Substitute:

\[ \begin{aligned} f'(x) &= \frac{(\sec x + 1)(\sec x \tan x) - (\sec x - 1)(\sec x \tan x)}{(\sec x + 1)^2} \end{aligned} \]

Step 4: Factor common term

\[ \begin{aligned} &= \frac{\sec x \tan x \left[(\sec x + 1) - (\sec x - 1)\right]}{(\sec x + 1)^2} \end{aligned} \]

Step 5: Simplify bracket

\[ \begin{aligned} (\sec x + 1) - (\sec x - 1) = 2 \end{aligned} \]

Thus,

\[ \begin{aligned} f'(x) &= \frac{2\sec x \tan x}{(\sec x + 1)^2} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{2\sec x \tan x}{(\sec x + 1)^2} \]

Key Insight

Since \(u'\) and \(v'\) are identical, factorization becomes very powerful and simplifies the expression immediately.

Quick Recognition Trick

If numerator and denominator differ only by sign (\(\pm 1\)), expect cancellation after factoring.

Common Mistake

Forgetting that both derivatives are same leads to unnecessary expansion and errors.

Exam Significance

Common CBSE board question
Appears in JEE as simplification-based MCQ
Tests pattern recognition in quotient rule
Improves algebraic efficiency in calculus

Concept Used

Chain Rule
Power Rule
Derivative: \(\dfrac{d}{dx}(\sin x)=\cos x\)

If \(y = [g(x)]^n\), then \[ \frac{dy}{dx} = n[g(x)]^{n-1}\cdot g'(x) \]

Solution Roadmap

Step 1: Identify inner function \(g(x)=\sin x\)
Step 2: Apply power rule
Step 3: Multiply by derivative of inner function
Step 4: Simplify

Solution

Power of sine function → derivative combines power rule and trig derivative

Step 1: Rewrite the function

\[ f(x) = (\sin x)^n \]

Step 2: Apply chain rule

\[ \begin{aligned} f'(x) &= n(\sin x)^{n-1} \cdot \frac{d}{dx}(\sin x) \end{aligned} \]

Step 3: Differentiate inner function

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

Step 4: Substitute

\[ \begin{aligned} f'(x) &= n(\sin x)^{n-1} \cdot \cos x \ &= n\sin^{\,n-1}x \cos x \end{aligned} \]

Final Answer:

\[ f'(x) = n\sin^{\,n-1}x \cos x \]

Key Insight

Treat \(\sin^n x\) as \((\sin x)^n\), not \(\sin(nx)\). This distinction is crucial in differentiation.

Quick Recognition Trick

For \((\sin x)^n\): multiply by \(n\), reduce power by 1, and multiply by \(\cos x\).

Common Mistake

Confusing \(\sin^n x\) with \(\sin(nx)\), or forgetting to multiply by \(\cos x\).

Exam Significance

Very common CBSE board question
Frequently appears in JEE/NEET MCQs
Core example of chain rule application
Foundation for higher trig power derivatives

Concept Used

Quotient Rule
Trigonometric derivatives
Identity: \(\sin^2 x + \cos^2 x = 1\)

\[ \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v u' - u v'}{v^2} \]

Solution Roadmap

Step 1: Identify numerator and denominator
Step 2: Differentiate each
Step 3: Apply quotient rule
Step 4: Expand numerator carefully
Step 5: Use identity to simplify

Solution

Step 1: Define functions

\[ u = a + b\sin x,\quad v = c + d\cos x \]

Step 2: Differentiate

\[ \begin{aligned} u' &= b\cos x \\ v' &= -d\sin x \end{aligned} \]

Step 3: Apply quotient rule

\[ f'(x) = \frac{v u' - u v'}{v^2} \]

Substitute values:

\[ \begin{aligned} f'(x) &= \frac{(c+d\cos x)(b\cos x) - (a+b\sin x)(-d\sin x)}{(c+d\cos x)^2} \end{aligned} \]

Step 4: Expand numerator

\[ \begin{aligned} (c+d\cos x)(b\cos x) &= bc\cos x + bd\cos^2 x \\ (a+b\sin x)(-d\sin x) &= -ad\sin x - bd\sin^2 x \end{aligned} \]

Since subtraction of a negative becomes addition:

\[ \begin{aligned} f'(x) &= \frac{bc\cos x + bd\cos^2 x + ad\sin x + bd\sin^2 x}{(c+d\cos x)^2} \end{aligned} \]

Step 5: Use identity

\[ \cos^2 x + \sin^2 x = 1 \]

\[ \begin{aligned} bd\cos^2 x + bd\sin^2 x = bd \end{aligned} \]

Thus,

\[ \begin{aligned} f'(x) &= \frac{bc\cos x + ad\sin x + bd}{(c+d\cos x)^2} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{bc\cos x + ad\sin x + bd}{(c+d\cos x)^2} \]

Key Insight

Grouping terms with \(\sin^2 x\) and \(\cos^2 x\) allows immediate simplification using identity, drastically reducing complexity.

Quick Recognition Trick

Whenever \(\sin^2 x\) and \(\cos^2 x\) appear with same coefficient, directly replace their sum with 1.

Common Mistake

Missing the sign change in \(-(a+b\sin x)(-d\sin x)\) or forgetting identity simplification.

Exam Significance

Very common CBSE board question
Frequently appears in JEE as simplification-heavy MCQ
Tests quotient rule + trig identities together
Improves algebraic manipulation speed

Concept Used

Constant Multiple Rule: \(\dfrac{d}{dx}[k f(x)] = k f'(x)\)
Chain Rule
\(\dfrac{d}{dx}(\sin x) = \cos x\)

Since \(\cos a\) is constant, it can be taken outside differentiation.

Solution Roadmap

Step 1: Treat denominator as constant
Step 2: Apply constant multiple rule
Step 3: Differentiate \(\sin(x+a)\) using chain rule
Step 4: Simplify

Solution

Step 1: Rewrite the function

\[ f(x) = \frac{1}{\cos a}\sin(x+a) \]

Step 2: Apply constant multiple rule

\[ f'(x) = \frac{1}{\cos a}\cdot \frac{d}{dx}\sin(x+a) \]

Step 3: Apply chain rule

\[ \frac{d}{dx}\sin(x+a) = \cos(x+a)\cdot \frac{d}{dx}(x+a) \]

\[ \frac{d}{dx}(x+a) = 1 \]

Thus,

\[ \frac{d}{dx}\sin(x+a) = \cos(x+a) \]

Step 4: Substitute

\[ \begin{aligned} f'(x) &= \frac{1}{\cos a}\cdot \cos(x+a) \\ &= \frac{\cos(x+a)}{\cos a} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{\cos(x+a)}{\cos a} \]

Key Insight

Constants (like \(\cos a\)) do not affect differentiation except as scaling factors.

Quick Recognition Trick

If denominator is constant, directly differentiate numerator and keep denominator unchanged.

Common Mistake

Treating \(\cos a\) as variable instead of constant, leading to incorrect quotient rule usage.

Exam Significance

Common CBSE board conceptual question
Tests understanding of constants vs variables
Appears in JEE/NEET as quick MCQ
Strengthens chain rule fundamentals

Concept Used

Product Rule: \(\dfrac{d}{dx}(uv)=u'v+uv'\)
Power Rule: \(\dfrac{d}{dx}(x^n)=nx^{n-1}\)
Trigonometric derivatives

It is often helpful to expand before differentiating.

Solution Roadmap

Step 1: Expand the expression
Step 2: Apply product rule to each term
Step 3: Simplify each derivative
Step 4: Combine and factor terms

Solution

Step 1: Expand the expression

\[ f(x) = x^4(5\sin x - 3\cos x) = 5x^4\sin x - 3x^4\cos x \]

Step 2: Differentiate first term

\[ \frac{d}{dx}(5x^4\sin x) = 5\frac{d}{dx}(x^4\sin x) \]

Apply product rule:

\[ \begin{aligned} \frac{d}{dx}(x^4\sin x) &= \sin x \cdot \frac{d}{dx}(x^4) + x^4 \cdot \frac{d}{dx}(\sin x) \\ &= \sin x (4x^3) + x^4(\cos x) \\ &= 4x^3\sin x + x^4\cos x \end{aligned} \]

Multiply by 5:

\[ = 20x^3\sin x + 5x^4\cos x \]

Step 3: Differentiate second term

\[ \frac{d}{dx}(3x^4\cos x) = 3\frac{d}{dx}(x^4\cos x) \]

Apply product rule:

\[ \begin{aligned} \frac{d}{dx}(x^4\cos x) &= \cos x (4x^3) + x^4(-\sin x) \\ &= 4x^3\cos x - x^4\sin x \end{aligned} \]

Multiply by 3:

\[ = 12x^3\cos x - 3x^4\sin x \]

Step 4: Combine results

\[ \begin{aligned} f'(x) &= 20x^3\sin x + 5x^4\cos x - (12x^3\cos x - 3x^4\sin x) \end{aligned} \]

Distribute negative sign:

\[ \begin{aligned} &= 20x^3\sin x + 5x^4\cos x - 12x^3\cos x + 3x^4\sin x \end{aligned} \]

Step 5: Group terms

\[ \begin{aligned} &= (20x^3\sin x - 12x^3\cos x) + (3x^4\sin x + 5x^4\cos x) \end{aligned} \]

Factor:

\[ \begin{aligned} &= x^3(20\sin x - 12\cos x) + x^4(3\sin x + 5\cos x) \end{aligned} \]

Final Answer:

\[ f'(x) = x^3(20\sin x - 12\cos x) + x^4(3\sin x + 5\cos x) \]

Key Insight

Expanding first reduces complexity. Otherwise, applying product rule directly becomes cumbersome.

Quick Recognition Trick

For \(x^n \cdot \text{trig}\), expect result of form:
\(x^{n-1}(\text{trig}) + x^n(\text{other trig})\)

Common Mistake

Missing negative sign in derivative of \(\cos x\), or forgetting to distribute minus properly.

Exam Significance

Classic CBSE board long-answer problem
Common in JEE for product rule mastery
Tests handling of polynomial × trig functions
Important for higher calculus applications

Concept Used

Product Rule: \(\dfrac{d}{dx}(uv)=u'v+uv'\)
Power Rule
Trigonometric derivative: \(\dfrac{d}{dx}(\cos x)=-\sin x\)

Solution Roadmap

Step 1: Identify \(u\) and \(v\)
Step 2: Differentiate each function
Step 3: Apply product rule
Step 4: Simplify and optionally rearrange

Solution

Step 1: Define functions

\[ u = x^2 + 1,\quad v = \cos x \]

Step 2: Differentiate

\[ u' = 2x,\quad v' = -\sin x \]

Step 3: Apply product rule

\[ \begin{aligned} f'(x) &= u'v + uv' \\ &= (2x)(\cos x) + (x^2+1)(-\sin x) \end{aligned} \]

Step 4: Simplify

\[ \begin{aligned} f'(x) &= 2x\cos x - (x^2+1)\sin x \end{aligned} \]

Step 5: Expand (optional form)

\[ \begin{aligned} &= 2x\cos x - x^2\sin x - \sin x \end{aligned} \]

Final Answer:

\[ f'(x) = 2x\cos x - (x^2+1)\sin x \]

Key Insight

Keeping the answer in factored form \((x^2+1)\sin x\) is often cleaner and preferred in exams.

Quick Recognition Trick

For polynomial × trig, derivative always follows pattern:
(derivative of polynomial × trig) + (polynomial × derivative of trig)

Common Mistake

Missing negative sign in derivative of \(\cos x\), or not applying product rule correctly.

Exam Significance

Very common CBSE board question
Frequently appears in JEE/NEET
Tests product rule fundamentals
Foundation for more complex mixed functions

Concept Used

Product Rule: \(\dfrac{d}{dx}(uv)=u'v+uv'\)
Power Rule
Trigonometric derivatives

Solution Roadmap

Step 1: Identify \(u\) and \(v\)
Step 2: Differentiate each function carefully
Step 3: Apply product rule
Step 4: Simplify and optionally expand

Solution

Step 1: Define functions

\[ u = ax^2 + \sin x,\quad v = p + q\cos x \]

Step 2: Differentiate each

\[ \begin{aligned} u' &= \frac{d}{dx}(ax^2) + \frac{d}{dx}(\sin x) \\ &= 2ax + \cos x \end{aligned} \]

\[ \begin{aligned} v' &= \frac{d}{dx}(p) + \frac{d}{dx}(q\cos x) \\ &= 0 + q(-\sin x) \\ &= -q\sin x \end{aligned} \]

Step 3: Apply product rule

\[ \begin{aligned} f'(x) &= u'v + uv' \\ &= (2ax + \cos x)(p + q\cos x) + (ax^2 + \sin x)(-q\sin x) \end{aligned} \]

Step 4: Simplify

\[ \begin{aligned} f'(x) &= (p + q\cos x)(2ax + \cos x) - (ax^2 + \sin x)(q\sin x) \end{aligned} \]

Optional Expansion (for clarity)

\[ \begin{aligned} (p + q\cos x)(2ax + \cos x) &= 2apx + p\cos x + 2aqx\cos x + q\cos^2 x \end{aligned} \]

\[ (ax^2 + \sin x)(q\sin x) = aqx^2\sin x + q\sin^2 x \]

So expanded form:

\[ \begin{aligned} f'(x) &= 2apx + p\cos x + 2aqx\cos x + q\cos^2 x \\ &\quad - aqx^2\sin x - q\sin^2 x \end{aligned} \]

Final Answer (preferred form):

\[ f'(x) = (p + q\cos x)(2ax + \cos x) - (ax^2 + \sin x)(q\sin x) \]

Key Insight

Keeping expressions in factored form avoids unnecessary expansion and reduces algebraic errors.

Quick Recognition Trick

Always compute \(u'\) and \(v'\) first, then directly plug into product rule without expanding early.

Common Mistake

Missing negative sign in derivative of \(\cos x\), or expanding too early leading to algebra errors.

Exam Significance

High-value CBSE board question
Common in JEE for multi-term product rule
Tests handling of algebra + trig together
Important for advanced calculus problems

Concept Used

Product Rule: \(\dfrac{d}{dx}(uv)=u'v+uv'\)
Trigonometric derivatives:

\[ \begin{aligned} \frac{d}{dx}(\cos x)=-\sin x,\\ \frac{d}{dx}(\tan x)=\sec^2 x \end{aligned} \]

Solution Roadmap

Step 1: Identify \(u\) and \(v\)
Step 2: Differentiate each function
Step 3: Apply product rule
Step 4: Simplify using identities

Solution

Step 1: Define functions

\[ \begin{aligned} u = x+\cos x,\\ v = x-\tan x \end{aligned} \]

Step 2: Differentiate each

\[ \begin{aligned} u' &= \frac{d}{dx}(x) + \frac{d}{dx}(\cos x) \\ &= 1 - \sin x \end{aligned} \]

\[ \begin{aligned} v' &= \frac{d}{dx}(x) - \frac{d}{dx}(\tan x) \\ &= 1 - \sec^2 x \end{aligned} \]

Step 3: Apply product rule

\[ \begin{aligned} f'(x) &= u'v + uv' \\ &= (1-\sin x)(x-\tan x) + (x+\cos x)(1-\sec^2 x) \end{aligned} \]

Step 4: Use identity

\[ \sec^2 x = 1 + \tan^2 x \]

\[ 1 - \sec^2 x = -\tan^2 x \]

Substitute:

\[ \begin{aligned} f'(x) &= (x-\tan x)(1-\sin x) + (x+\cos x)(-\tan^2 x) \end{aligned} \]

Step 5: Simplify

\[ \begin{aligned} f'(x) &= (x-\tan x)(1-\sin x) - \tan^2 x (x+\cos x) \end{aligned} \]

Final Answer:

\[ f'(x) = (x-\tan x)(1-\sin x) - \tan^2 x (x+\cos x) \]

Key Insight

Recognizing \(1-\sec^2 x = -\tan^2 x\) avoids unnecessary expansion and simplifies the result quickly.

Quick Recognition Trick

Always replace \(1-\sec^2 x\) immediately to simplify expressions involving \(\tan x\).

Common Mistake

Forgetting derivative of \(\tan x = \sec^2 x\), or missing identity transformation step.

Exam Significance

Important CBSE board question
Frequently appears in JEE for identity simplification
Tests product rule + trig identity integration
Enhances speed and algebraic control

Concept Used

Quotient Rule: \(\dfrac{d}{dx}\left(\frac{u}{v}\right)=\frac{v u' - u v'}{v^2}\)
Trigonometric derivatives
Identity: \(\sin^2 x + \cos^2 x = 1\)

Solution Roadmap

Step 1: Identify \(u\) and \(v\)
Step 2: Differentiate each
Step 3: Apply quotient rule
Step 4: Expand numerator carefully
Step 5: Use identity to simplify

Solution

Step 1: Define functions

\[ \begin{aligned} u = 4x + 5\sin x,\\ v = 3x + 7\cos x \end{aligned} \]

Step 2: Differentiate

\[ \begin{aligned} u' &= 4 + 5\cos x \\ v' &= 3 - 7\sin x \end{aligned} \]

Step 3: Apply quotient rule

\[ f'(x) = \frac{v u' - u v'}{v^2} \]

Substitute:

\[ \begin{aligned} f'(x) &= \frac{(3x+7\cos x)(4+5\cos x) - (4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^2} \end{aligned} \]

Step 4: Expand first product

\[ \begin{aligned} (3x+7\cos x)(4+5\cos x) &= 12x + 15x\cos x + 28\cos x + 35\cos^2 x \end{aligned} \]

Step 5: Expand second product

\[ \begin{aligned} (4x+5\sin x)(3-7\sin x) &= 12x - 28x\sin x + 15\sin x - 35\sin^2 x \end{aligned} \]

Step 6: Subtract carefully

\[ \begin{aligned} f'(x) &= \frac{[12x + 15x\cos x + 28\cos x + 35\cos^2 x] - [12x - 28x\sin x + 15\sin x - 35\sin^2 x]}{(3x+7\cos x)^2} \end{aligned} \]

Distribute minus:

\[ \begin{aligned} &= \frac{12x + 15x\cos x + 28\cos x + 35\cos^2 x - 12x + 28x\sin x - 15\sin x + 35\sin^2 x}{(3x+7\cos x)^2} \end{aligned} \]

Step 7: Simplify

\[ \begin{aligned} &= \frac{15x\cos x + 28x\sin x + 28\cos x - 15\sin x + 35(\cos^2 x + \sin^2 x)}{(3x+7\cos x)^2} \end{aligned} \]

Use identity:

\[ \cos^2 x + \sin^2 x = 1 \]

\[ \begin{aligned} f'(x) &= \frac{15x\cos x + 28x\sin x + 28\cos x - 15\sin x + 35}{(3x+7\cos x)^2} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{15x\cos x + 28x\sin x + 28\cos x - 15\sin x + 35}{(3x+7\cos x)^2} \]

Key Insight

Always combine \(\cos^2 x\) and \(\sin^2 x\) early — it drastically simplifies large expressions.

Quick Recognition Trick

After expansion, scan immediately for \(\sin^2 x + \cos^2 x\) to reduce complexity.

Common Mistake

Sign errors during subtraction and forgetting identity simplification.

Exam Significance

Very important CBSE board long-answer question
Common in JEE for algebra-heavy differentiation
Tests quotient rule mastery
Improves accuracy under complex expansions

Concept Used

Constant Multiple Rule
Quotient Rule
Power Rule
Trigonometric derivatives

\[ \cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \quad (\text{constant}) \]

Solution Roadmap

Step 1: Take constant outside
Step 2: Apply quotient rule to remaining function
Step 3: Differentiate numerator and denominator
Step 4: Simplify and factor

Solution

Step 1: Extract constant

\[ f(x) = \cos\left(\frac{\pi}{4}\right)\cdot \frac{x^2}{\sin x} \]

Step 2: Apply quotient rule to \(\frac{x^2}{\sin x}\)

Let \(u = x^2,\; v = \sin x\)

\[ \begin{aligned} u' = 2x,\\ v' = \cos x \end{aligned} \]

\[ \begin{aligned} \frac{d}{dx}\left(\frac{x^2}{\sin x}\right) &= \frac{\sin x(2x) - x^2(\cos x)}{\sin^2 x} \end{aligned} \]

Step 3: Multiply by constant

\[ \begin{aligned} f'(x) &= \cos\left(\frac{\pi}{4}\right)\cdot \frac{2x\sin x - x^2\cos x}{\sin^2 x} \end{aligned} \]

Step 4: Write final form

\[ f'(x) = \frac{\cos\left(\frac{\pi}{4}\right)\left(2x\sin x - x^2\cos x\right)}{\sin^2 x} \]

Optional Simplification

\[ \cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \]

\[ f'(x)=\frac{1}{\sqrt{2}}\cdot \frac{2x\sin x - x^2\cos x}{\sin^2 x} \]

Key Insight

Always extract constants first — it simplifies the structure and reduces computation errors.

Quick Recognition Trick

Treat \(\cos(\pi/4)\) as constant immediately; avoid applying quotient rule on entire expression.

Common Mistake

Not recognizing \(\cos(\pi/4)\) as constant, leading to unnecessary complexity.

Exam Significance

Common CBSE board conceptual question
Tests ability to identify constants
Appears in JEE as simplification-based MCQ
Improves efficiency in quotient rule problems

Concept Used

Quotient Rule
Derivative: \(\dfrac{d}{dx}(\tan x)=\sec^2 x\)
Identity awareness: \(\sec^2 x = 1+\tan^2 x\)

\[ \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v u' - u v'}{v^2} \]

Solution Roadmap

Step 1: Identify numerator and denominator
Step 2: Differentiate each
Step 3: Apply quotient rule
Step 4: Simplify expression

Solution

Step 1: Define functions

\[ u = x,\quad v = 1+\tan x \]

Step 2: Differentiate

\[ \begin{aligned} u' = 1,\\ v' = \sec^2 x \end{aligned} \]

Step 3: Apply quotient rule

\[ \begin{aligned} f'(x) &= \frac{v u' - u v'}{v^2} \\ &= \frac{(1+\tan x)(1) - x(\sec^2 x)}{(1+\tan x)^2} \end{aligned} \]

Step 4: Simplify

\[ \begin{aligned} f'(x) &= \frac{1+\tan x - x\sec^2 x}{(1+\tan x)^2} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{1+\tan x - x\sec^2 x}{(1+\tan x)^2} \]

Optional Alternate Form

Using \(\sec^2 x = 1+\tan^2 x\),

\[ f'(x) = \frac{1+\tan x - x(1+\tan^2 x)}{(1+\tan x)^2} \]

Key Insight

Quotient rule problems involving \(\tan x\) almost always bring \(\sec^2 x\), so be ready with identities.

Quick Recognition Trick

Always compute numerator and denominator derivatives first, then directly substitute to avoid confusion.

Common Mistake

Forgetting derivative of \(\tan x = \sec^2 x\) or missing square in denominator.

Exam Significance

Common CBSE board question
Appears in JEE for quotient rule practice
Tests trig derivative memory
Improves speed in mixed algebra-trig problems

Concept Used

Product Rule: \(\dfrac{d}{dx}(uv)=u'v+uv'\)
Trigonometric derivatives:

\[ \frac{d}{dx}(\sec x)=\sec x \tan x,\quad \frac{d}{dx}(\tan x)=\sec^2 x \]

Solution Roadmap

Step 1: Identify \(u\) and \(v\)
Step 2: Differentiate each function
Step 3: Apply product rule
Step 4: Use identities to simplify

Solution

Step 1: Define functions

\[ u = x+\sec x,\quad v = x-\tan x \]

Step 2: Differentiate

\[ \begin{aligned} u' &= \frac{d}{dx}(x) + \frac{d}{dx}(\sec x) \\ &= 1 + \sec x \tan x \end{aligned} \]

\[ \begin{aligned} v' &= \frac{d}{dx}(x) - \frac{d}{dx}(\tan x) \\ &= 1 - \sec^2 x \end{aligned} \]

Step 3: Apply product rule

\[ \begin{aligned} f'(x) &= u'v + uv' \\ &= (1+\sec x\tan x)(x-\tan x) + (x+\sec x)(1-\sec^2 x) \end{aligned} \]

Step 4: Use identity

\[ \sec^2 x = 1 + \tan^2 x \Rightarrow 1 - \sec^2 x = -\tan^2 x \]

Substitute:

\[ \begin{aligned} f'(x) &= (x-\tan x)(1+\sec x\tan x) - \tan^2 x (x+\sec x) \end{aligned} \]

Final Answer:

\[ f'(x) = (x-\tan x)(1+\sec x\tan x) - \tan^2 x (x+\sec x) \]

Optional Expansion (for deeper clarity)

Expanding is possible but not recommended in exams due to complexity.

Key Insight

Recognizing \(1-\sec^2 x = -\tan^2 x\) is the key simplification step in sec–tan problems.

Quick Recognition Trick

Always convert \(1-\sec^2 x\) immediately — it avoids unnecessary expansion.

Common Mistake

Confusing derivative of \(\sec x\) or forgetting identity \(\sec^2 x = 1+\tan^2 x\).

Exam Significance

Final-level CBSE board question
Common in JEE for mixed trig structures
Tests multi-rule integration
Strengthens identity-based simplification

Concept Used

Quotient Rule
Chain Rule for \((\sin x)^n\)
Power Rule

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

Solution Roadmap

Step 1: Identify numerator and denominator
Step 2: Differentiate each part
Step 3: Apply quotient rule
Step 4: Factor common powers of \(\sin x\)
Step 5: Simplify final expression

Solution

Step 1: Define functions

\[ \begin{aligned} u &= x,\\ v &= \sin^n x \end{aligned} \]

Step 2: Differentiate

\[ u' = 1 \]

\[ \begin{aligned} v' &= \frac{d}{dx}(\sin^n x) \\&= n\sin^{n-1}x \cos x \end{aligned} \]

Step 3: Apply quotient rule

\[ \begin{aligned} f'(x) &= \frac{v u' - u v'}{v^2} \\ &= \frac{\sin^n x(1) - x(n\sin^{n-1}x\cos x)}{\sin^{2n}x} \end{aligned} \]

Step 4: Factor common term

\[ \begin{aligned} &= \frac{\sin^{n-1}x\left(\sin x - nx\cos x\right)}{\sin^{2n}x} \end{aligned} \]

Step 5: Simplify powers

\[ \sin^{2n}x = \sin^{n-1}x \cdot \sin^{n+1}x \]

\[ \begin{aligned} f'(x) &= \frac{\sin x - nx\cos x}{\sin^{n+1}x} \end{aligned} \]

Final Answer:

\[ f'(x) = \frac{\sin x - nx\cos x}{\sin^{n+1}x} \]

Key Insight

Factoring \(\sin^{n-1}x\) early simplifies the expression dramatically and avoids messy exponents.

Quick Recognition Trick

For \(\frac{x}{\sin^n x}\), expect final denominator to increase power → \(\sin^{n+1}x\).

Common Mistake

Forgetting chain rule in \(\sin^n x\) or mishandling powers during simplification.

Exam Significance

Very important CBSE board question
Frequently appears in JEE as power + quotient combination
Tests multi-rule integration
Builds strong algebraic simplification skills

Chapter 12 — LIMITS AND DERIVATIVES

Concept Used

Solution Roadmap

Solution

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Common Mistake

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Common Mistake

Exam Significance

Concept Used

Solution Roadmap

Solution

Alternative Method (Quotient Rule)

Key Insight

Common Mistake

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Common Mistake

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Common Mistake

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Common Mistake

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Common Mistake

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Quick Recognition Trick

Common Mistake

Exam Significance

Concept Used

Solution Roadmap

Solution

Key Insight

Quick Recognition Trick

Common Mistake

Exam Significance

Concept Used

Solution Roadmap

Solution