यदि $f(x) = \cos^{-1} \left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right]$ है,तो $f'(e) = \_\_\_\_$

  • A
    $1/e$
  • B
    $2/e^2$
  • C
    $2/e$
  • D
    $1$

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Similar Questions

यदि $y(x) = \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right)$,जहाँ $x \in \left(\frac{\pi}{2}, \pi\right)$,तो $x = \frac{5\pi}{6}$ पर $\frac{dy}{dx}$ का मान ज्ञात कीजिए।

यदि $y = \tan^{-1} \left( \frac{3\cos x - 4\sin x}{4\cos x + 3\sin x} \right) + 2\tan^{-1} \left( \frac{x}{1+\sqrt{1-x^2}} \right)$ है,तो $x = \frac{\sqrt{3}}{2}$ पर $\frac{dy}{dx}$ का मान ज्ञात कीजिए:

यदि $y = \sin^{-1} \left[ \frac{\sqrt{1+x} + \sqrt{1-x}}{2} \right]$ है,तो $\frac{dy}{dx} = $

यदि $y = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$,जहाँ $0 \leqslant x < \frac{\pi}{2}$,तो $y'\left(\frac{\pi}{6}\right)$ का मान ज्ञात कीजिए।

यदि $y = \operatorname{Tan}^{-1}\left(\frac{2x}{1-x^2}\right)$ जहाँ $|x| < 1$,तो $x = \frac{1}{2}$ पर $\left(\frac{dy}{dx}\right)$ का मान ज्ञात कीजिए।

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