यदि $y = \tan^{-1}\left(\frac{12x - 64x^3}{1 - 48x^2}\right)$ है,तो $\frac{dy}{dx} = $

  • A
    $\frac{3}{1 + 16x^2}$
  • B
    $\frac{4}{1 + 16x^2}$
  • C
    $\frac{12}{1 + 16x^2}$
  • D
    $\frac{1}{1 + 16x^2}$

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यदि $f(x)=\cos ^{-1}\left[\frac{1}{\sqrt{13}}(2 \cos x-3 \sin x)\right]$ है,तो $f^{\prime}(0.5)$ का मान ज्ञात कीजिए।

$y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)$ का अवकलज क्या है?

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 $0 < |x| < 1$ के लिए $y = \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$ है,तो $\frac{dy}{dx} = $

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