JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

f(x)=\cos ^{-1}\left\{\frac{1-\left(\log _e x\right)^2}{1+\left(\log _e x\right)^2}\right\}

, then

f^{\prime}\left(\frac{1}{e}\right)

is equal to

View answer and explanation

Correct option: B

-e

Option B is correct because it correctly simplifies the inverse cosine function and finds its derivative, then evaluates it at the given point.

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