CUET UG Mathematics Calculus Differentiation Chain Rule Implicit Logarithmic Differentiation

Must-Know

• The chain rule states that if ( y = f(u) ) and ( u = g(x) ), then ( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ). Example: ( y = \sin(x^2) ), ( \frac{dy}{dx} = \cos(x^2) \cdot 2x ).
• For composite functions like ( y = (3x + 2)^5 ), apply chain rule: ( \frac{dy}{dx} = 5(3x+2)^4 \cdot 3 = 15(3x+2)^4 ).
• Derivative of ( e^{f(x)} ) is ( e^{f(x)} \cdot f'(x) ). Example: ( \frac{d}{dx}(e^{\sin x}) = e^{\sin x} \cdot \cos x ).
• Derivative of ( \ln(f(x)) ) is ( \frac{1}{f(x)} \cdot f'(x) ), valid only if ( f(x) > 0 ). Example: ( \frac{d}{dx}[\ln(x^2 + 1)] = \frac{2x}{x^2 + 1} ).
• Logarithmic differentiation is used when function is of the form ( [f(x)]^{g(x)} ) or product/quotient of multiple functions. Take ( \ln ) on both sides first.
• To differentiate ( y = x^x ), take ( \ln y = x \ln x ), then differentiate: ( \frac{1}{y} \frac{dy}{dx} = \ln x + 1 ), so ( \frac{dy}{dx} = x^x (\ln x + 1) ).
• For implicit functions like ( x^2 + y^2 = 25 ), differentiate both sides w.r.t. ( x ): ( 2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y} ).
• In implicit differentiation, treat ( y ) as a function of ( x ), so ( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} ).
• If ( y = \log_a(f(x)) ), then ( \frac{dy}{dx} = \frac{1}{f(x) \ln a} \cdot f'(x) ). Example: ( \frac{d}{dx}(\log_{10} x) = \frac{1}{x \ln 10} ).
• Chain rule applies to trigonometric compositions: ( \frac{d}{dx}[\cos(2x)] = -\sin(2x) \cdot 2 = -2\sin(2x) ).
• For ( y = \tan(\sqrt{x}) ), ( \frac{dy}{dx} = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} ).
• When differentiating inverse trig functions with chain rule: ( \frac{d}{dx}[\sin^{-1}(2x)] = \frac{2}{\sqrt{1 - (2x)^2}} = \frac{2}{\sqrt{1 - 4x^2}} ), valid for ( |2x| < 1 ).
• For ( y = \ln|\sin x| ), derivative is ( \frac{1}{\sin x} \cdot \cos x = \cot x ), defined where ( \sin x \ne 0 ).
• If ( y = \sqrt{\sin x} ), then ( \frac{dy}{dx} = \frac{1}{2\sqrt{\sin x}} \cdot \cos x = \frac{\cos x}{2\sqrt{\sin x}} ).
• For ( y = (x+1)^3(x+2)^4 ), use logarithmic differentiation: ( \ln y = 3\ln(x+1) + 4\ln(x+2) ), then ( \frac{y'}{y} = \frac{3}{x+1} + \frac{4}{x+2} ).
• The derivative of ( a^{f(x)} ) is ( a^{f(x)} \ln a \cdot f'(x) ). Example: ( \frac{d}{dx}(2^x) = 2^x \ln 2 ).
• For ( x^y = y^x ), apply logarithmic differentiation: take ( \ln ) both sides → ( y \ln x = x \ln y ), then differentiate implicitly.
• If ( y = \sin^{-1}(f(x)) ), then ( \frac{dy}{dx} = \frac{f'(x)}{\sqrt{1 - [f(x)]^2}} ), provided ( -1 < f(x) < 1 ).
• For ( y = \log(\log x) ), derivative is ( \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} ), valid for ( x > 1 ).
• Verify from NCERT: The derivative of ( \sec^{-1}x ) is ( \frac{1}{|x|\sqrt{x^2 - 1}} ), valid for ( |x| > 1 ).

Difficulty Level

Intermediate — Requires understanding of function composition and manipulation of logarithms, but avoids high-level proofs or multivariable concepts.

Common CUET Traps

• Trap: Forgetting to multiply by the derivative of the inner function in chain rule.
  Avoid: Always apply ( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ); never skip the ( \frac{du}{dx} ) part.
• Trap: Differentiating ( y^2 ) as ( 2y ) instead of ( 2y \frac{dy}{dx} ) in implicit functions.
  Avoid: Remember ( y ) is a function of ( x ), so use chain rule: ( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} ).
• Trap: Applying logarithmic rules to negative functions, e.g., ( \ln(-x) ) without checking domain.
  Avoid: ( \ln(f(x)) ) is defined only when ( f(x) > 0 ); always verify domain before differentiating.

Practice MCQs

1. Question: The derivative of ( y = \sin(3x + 2) ) with respect to ( x ) is:
   
   A. ( \cos(3x + 2) )
   
   B. ( 3\cos(3x + 2) )
   
   C. ( -\cos(3x + 2) )
   
   D. ( \sin(3) )
   Answer: B
   Explanation: By chain rule, derivative of ( \sin(u) ) is ( \cos(u) \cdot u' ), where ( u = 3x+2 ), so ( u' = 3 ).
   Why others fail: Option A misses multiplying by derivative of inner function (3).

2. Question: If ( y = \ln(x^3) ), then ( \frac{dy}{dx} ) is:
   
   A. ( \frac{3}{x} )
   
   B. ( \frac{1}{3x} )
   
   C. ( \frac{3}{x^3} )
   
   D. ( \frac{1}{x} )
   Answer: A
   Explanation: ( \ln(x^3) = 3\ln x ), so derivative is ( 3 \cdot \frac{1}{x} = \frac{3}{x} ).
   Why others fail: Option D ignores the exponent and treats it as ( \ln x ).

3. Question: For the implicit function ( x^2 + y^2 = 9 ), the value of ( \frac{dy}{dx} ) at ( (0,3) ) is:
   
   A. 0
   
   B. 1
   
   C. undefined
   
   D. –1
   Answer: A
   Explanation: Differentiate: ( 2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y} ); at ( (0,3) ), it is ( -\frac{0}{3} = 0 ).
   Why others fail: Option C arises from misreading denominator as zero, but ( y = 3 \ne 0 ).

4. Question: The derivative of ( y = x^{\sin x} ) is best found using:
   
   A. Product rule
   
   B. Chain rule only
   
   C. Logarithmic differentiation
   
   D. Quotient rule
   Answer: C
   Explanation: Since the function has variable base and exponent, logarithmic differentiation is required.
   Why others fail: Option B is tempting because chain rule is involved, but alone it's insufficient.

5. Question: If ( y = \log_{10}(e^x) ), then ( \frac{dy}{dx} ) is:
   
   A. ( \frac{1}{\ln 10} )
   
   B. ( \frac{e^x}{\ln 10} )
   
   C. ( \frac{1}{e^x \ln 10} )
   
   D. ( \ln 10 )
   Answer: A
   Explanation: ( \log_{10}(e^x) = x \log_{10} e = x \cdot \frac{1}{\ln 10} ), so derivative is ( \frac{1}{\ln 10} ).
   Why others fail: Option B incorrectly applies chain rule without simplifying the logarithmic expression first.

Last‑Minute Revision

• ⚠️ Chain rule: ( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ) — never skip inner derivative.
• ⚠️ ( \frac{d}{dx}(e^{f(x)}) = e^{f(x)} f'(x) ) — base ( e ) stays, multiply by derivative of exponent.
• ⚠️ ( \frac{d}{dx}[\ln(f(x))] = \frac{f'(x)}{f(x)} ) — only if ( f(x) > 0 ).
• ⚠️ Logarithmic differentiation: use for ( y = [f(x)]^{g(x)} ) or complex products.
• ⚠️ After taking ( \ln ) both sides, differentiate implicitly.
• ⚠️ ( \frac{d}{dx}(a^{f(x)}) = a^{f(x)} \ln a \cdot f'(x) ) — don’t forget ( \ln a ).
• ⚠️ For ( x^2 + y^2 = r^2 ), ( \frac{dy}{dx} = -\frac{x}{y} ) — standard result for circle.
• ⚠️ Treat ( y ) as function of ( x ) in implicit diff: ( \frac{d}{dx}(y^n) = n y^{n-1} \frac{dy}{dx} ).
• ⚠️ ( \frac{d}{dx}(\sin^{-1}u) = \frac{u'}{\sqrt{1 - u^2}} ) — domain: ( |u| < 1 ).
• ⚠️ ( \frac{d}{dx}(\cos^{-1}u) = -\frac{u'}{\sqrt{1 - u^2}} ) — negative sign is crucial.
• ⚠️ ( \frac{d}{dx}(\tan^{-1}u) = \frac{u'}{1 + u^2} ) — denominator is ( 1 + u^2 ), not minus.
• ⚠️ ( \frac{d}{dx}(\sec^{-1}x) = \frac{1}{|x|\sqrt{x^2 - 1}} ), ( |x| > 1 ).
• ⚠️ ( \frac{d}{dx}(\csc^{-1}x) = -\frac{1}{|x|\sqrt{x^2 - 1}} ), ( |x| > 1 ).
• ⚠️ For ( y = \ln|f(x)| ), derivative is ( \frac{f'(x)}{f(x)} ), same as without modulus.
• ⚠️ ( \frac{d}{dx}(x^x) = x^x(1 + \ln x) ) — derived via logarithmic differentiation.
• ⚠️ Never differentiate ( \ln(-x) ) directly; use ( \ln|x| ) and domain restrictions.
• ⚠️ Verify from NCERT: Derivatives of inverse trig functions include domain conditions.
• ⚠️ Mnemonic: "Log first, then diff" — for logarithmic differentiation, always take log before differentiating.