જો $y=\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}$ હોય,તો $x=0$ આગળ $\left(\frac{dy}{dx}\right)$ ની કિંમત શોધો.

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

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Similar Questions

જો $y = \tan^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}\right)$ હોય,તો $\frac{dy}{dx}$ ની કિંમત શોધો.

$\frac{d}{{dy}}\left( {{{\sin }^{ - 1}}\left( {\frac{{3y}}{2} - \frac{{{y^3}}}{2}} \right)} \right) = $

જો $f(x) = \tan^{-1}\left(\frac{1}{\sin^2 x + \sin x + 1}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 3\sin x + 3}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 5\sin x + 7}\right) + \dots$ $10$ પદો સુધી હોય, તો $f'(0) = $

જો $a > b > 0$ અને $x$ લઘુકોણ હોય,તો $\frac{d}{dx} \left[ \cos^{-1} \left( \frac{b - a \cos x}{a - b \cos x} \right) \right] = $

જો $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$ હોય,તો $\frac{d y}{d x}=$

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