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Mathematics Question on Continuity and differentiability

If f(x)=log⁡x2(log⁡ x)f(x) = \log_{x^2} (\log \,x), then f′(x)f'(x) at x=ex = e is

A

00

B

11

C

1e\frac{1}{e}

D

12e\frac{1}{ 2e}

Answer

12e\frac{1}{ 2e}

Explanation

Solution

f(x)=log⁡x2(log⁡x)f\left(x\right) = \log_{x^{2}}\left(\log x\right)
⇒  f(x)=12[log⁡x(log⁡x)][∵  log⁡anb=1nlog⁡ab]\Rightarrow \:\: f\left(x\right) =\frac{1}{2} \left[\log_{x}\left(\log x\right) \right] \left[\because \:\: \log_{a^{n}} b=\frac{1}{n} \log_{a^{b}} \right]
⇒f(x)=12[log⁡(log⁡x)log⁡x](∵  log⁡ab=log⁡blog⁡a)\Rightarrow f\left(x\right) =\frac{1}{2} \left[\frac{\log \left(\log x\right) }{\log x}\right] \left(\because \:\: \log_{a} b= \frac{\log b}{\log a } \right)
⇒f′(x)=12[log⁡(x)ddx[log⁡(log⁡(x))]−log⁡(log⁡x)ddxlog⁡x(log⁡x)2]\Rightarrow f'\left(x\right) =\frac{1}{2} \left[\frac{\log\left(x\right) \frac{d}{dx}\left[\log \left(\log \left(x\right)\right)\right]-\log \left(\log x\right) \frac{d}{dx}\log x}{\left(\log x\right)^{2}}\right]
=12[log⁡(x)1log⁡x×1x−log⁡(log⁡x)1x(log⁡x)2]=\frac{1}{2} \left[\frac{\log\left(x\right) \frac{1}{\log x}\times\frac{1}{x} -\log\left(\log x\right) \frac{1}{x}}{\left(\log x\right)^{2}}\right]
=12[1x−log⁡(log⁡(x))x(log⁡x)2]=\frac{1}{2}\left[\frac{\frac{1}{x} -\frac{\log\left(\log \left(x\right)\right)}{x}}{\left(\log x\right)^{2}}\right]
⇒f′(x)=12[1−log⁡(log⁡(x))x × (log⁡x)2]\Rightarrow f'\left(x\right) =\frac{1}{2}\left[\frac{1 -\log\left(\log \left(x\right)\right)}{x \, \times \,\left(\log x\right)^{2}}\right]
⇒f′(e)=12[1−log⁡(log⁡(e))e(log⁡e)2]\Rightarrow f'\left(e\right) =\frac{1}{2}\left[\frac{1 -\log\left(\log \left(e\right)\right)}{e \left(\log e\right)^{2}}\right]
=12[1−0e]=12e= \frac{1}{2} \left[ \frac{1 - 0}{e} \right] = \frac{1}{2e}