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Question

If $$f(x) = \displaystyle \cos ^{-1}\left ( \frac{1-\left ( \log x \right )^{2}}{1+\left ( \log x \right )^{2}} \right )$$ then $$f'(e)$$ is equal to-


A
does not exist
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B
2e
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C
1e
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D
1
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Solution

The correct option is C $$\displaystyle \frac{1}{e}$$
Given : $$\displaystyle f\left ( x \right )=\cos^{-1}\left [ \dfrac{1-\left ( \log x \right )^{2}}{1+\left ( \log x \right )^{2}} \right ]$$
$$\Rightarrow f\left ( x \right )=2\tan^{-1}\left ( \log x \right )$$ ....... $$\left[\because \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2\tan^{-1} x\right]$$

$$f'\left ( x \right )=2\dfrac{1}{1+\left ( \log x \right )^{2}}.\dfrac{1}{x}$$

$$=\dfrac{2}{(1+\left ( \log x \right )^{2})x}$$


$$\therefore f'(e)=\dfrac{2}{(1+\left ( \log e \right )^{2})e}$$


$$=\dfrac{2}{(1+1))e}=\dfrac{2}{2e}=\dfrac{1}{e}$$ ..... $$[\because \log e=1]$$
$$\therefore f'\left ( e \right )=\dfrac{1}{e}$$

Mathematics

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