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If f(x)=\cos ^{-1}\left[\frac{1-(\log x)^{2}}{1+(\log x)^{2}}\right],  then the value of f^{\prime}(e)=

Option: 1

1


Option: 2

\\ {1 / \mathrm{e}}


Option: 3

{2 / \mathrm{e}} \\


Option: 4

{\frac{2}{e^{2}}}


Answers (1)

best_answer

 

 

Differentiation of Inverse Function -

Differentiation of Inverse Function

Let f(x) be a function that is both invertible and differentiable.  Let y = f −1 (x) be the inverse of f(x). Then,\frac{d y}{d x}=\frac{d}{d x}\left(f^{-1}(x)\right)=\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}

f^{\prime}\left(f^{-1}(x)\right) \neq 0,

Alternatively, if y = g(x) is the inverse of f(x), then

g(x)=\frac{1}{f^{\prime}(g(x))}

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f(x)=\cos ^{-1}\left[\frac{1-(\log x)^{2}}{1+(\log x)^{2}}\right]=2 \tan ^{-1}(\log x)\\ f^{\prime}(x)=2 \cdot \frac{1}{1+(\log x)^{2}} \cdot \frac{1}{x} \\ f^{\prime}(e)=\frac{1}{e}

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manish

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