यदि $y = \sin^{-1} \left( \frac{25 - x^2}{25 + x^2} \right)$ है, तो $y'(1)$ का मान ज्ञात कीजिए।

  • A
    $-\frac{5}{13}$
  • B
    $\frac{5}{13}$
  • C
    $\frac{13}{5}$
  • D
    $\frac{2}{7}$

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Similar Questions

जब $x \in \left( {0, \frac{\pi }{2}} \right)$ है,तो $\frac{x}{2}$ के सापेक्ष ${\tan ^{ - 1}}\left( {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$ का अवकलज क्या है?

यदि $y = \operatorname{Tan}^{-1}\left(\frac{x}{1+2x^2}\right) + \operatorname{Tan}^{-1}\left(\frac{x}{1+6x^2}\right)$ है,तो $\frac{dy}{dx} = $

${\tan ^{ - 1}}\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} $ का ${\cos ^{ - 1}}({x^2})$ के सापेक्ष अवकल गुणांक ज्ञात कीजिए।

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

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \right] = $

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