Method of DifferentiationmediumFree

Method of Differentiation: Question

JEE Maths question with a full step-by-step solution.

Question
If f(x)=log⁡x(log⁡x)f(x)=\log_{x}(\log x), then f′(x)f'(x) at x=ex=e is
A1e\dfrac{1}{e}correct
B11
Cee
DNone of these
Solution
Step 1: Rewrite f(x)f(x) using the change of base formula (all logs are natural logs):
f(x)=log⁡x(log⁡x)=log⁡(log⁡x)log⁡xf(x)=\log_{x}(\log x)=\dfrac{\log(\log x)}{\log x}
 Step 2: Differentiate using the quotient rule. Let u=log⁡(log⁡x)u=\log(\log x) and v=log⁡xv=\log x:
u′=1log⁡x⋅1x=1xlog⁡x,v′=1xu'=\dfrac{1}{\log x}\cdot\dfrac{1}{x}=\dfrac{1}{x\log x},\qquad v'=\dfrac{1}{x}
f′(x)=u′v−uv′v2=1xlog⁡x⋅log⁡x  −  log⁡(log⁡x)⋅1x(log⁡x)2=1−log⁡(log⁡x)x(log⁡x)2f'(x)=\dfrac{u'v-uv'}{v^{2}}=\dfrac{\dfrac{1}{x\log x}\cdot\log x\;-\;\log(\log x)\cdot\dfrac{1}{x}}{(\log x)^{2}}=\dfrac{1-\log(\log x)}{x(\log x)^{2}}
 Step 3: Evaluate at x=ex=e. Since log⁡e=1\log e=1 and log⁡(log⁡e)=log⁡1=0\log(\log e)=\log 1=0:
f′(e)=1−0e⋅12=1ef'(e)=\dfrac{1-0}{e\cdot 1^{2}}=\dfrac{1}{e}
 Correct answer: (1)
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