यदि $y=\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}$ है,तो $x=0$ पर $\left(\frac{dy}{dx}\right)$ का मान ज्ञात कीजिए।

  • A
    $1$
  • B
    $2$
  • C
    $-2$
  • D
    $-1$

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Similar Questions

$\tan ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)$ का $\cos ^{-1} x^2$ के सापेक्ष अवकलज क्या है?

यदि $y=\cos ^{-1}\left(\frac{a^2}{\sqrt{x^4+a^4}}\right)$ है,तो $\frac{d y}{d x}$ क्या है?

यदि $f(x) = \tan^{-1}\left(\frac{1}{\sin^2 x + \sin x + 1}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 3\sin x + 3}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 5\sin x + 7}\right) + \dots$ $10$ पदों तक है, तो $f'(0) = $

$\frac{d}{d x}\left(\cos ^{-1}\left(\frac{4 x^3}{27}-x\right)\right)=$

यदि $2y = {\left( {{{\cot }^{ - 1}}\left( {\frac{{\sqrt 3 \cos x + \sin x}}{{\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$ और $x \in \left( {0,\frac{\pi }{2}} \right)$ है,तो $\frac{{dy}}{{dx}}$ का मान ज्ञात कीजिए।

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