$\tan^{-1}\left( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right)$ નો વિકલન સહગુણક શોધો.

  • A
    $\sqrt{1-x^2}$
  • B
    $\frac{1}{\sqrt{1-x^2}}$
  • C
    $\frac{1}{2\sqrt{1-x^2}}$
  • D
    $x$

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જો $y = \tan^{-1} \left( \frac{3\cos x - 4\sin x}{4\cos x + 3\sin x} \right) + 2\tan^{-1} \left( \frac{x}{1+\sqrt{1-x^2}} \right)$ હોય, તો $x = \frac{\sqrt{3}}{2}$ આગળ $\frac{dy}{dx}$ ની કિંમત શોધો:

$\frac{d}{d x}\left[\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]$ ની કિંમત શોધો.

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

જો $\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = a(x - y)$ હોય,તો $\frac{dy}{dx} = $

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જ્યારે $x \in \left( {0, \frac{\pi }{2}} \right)$ હોય,ત્યારે $\frac{x}{2}$ ની સાપેક્ષમાં ${\tan ^{ - 1}}\left( {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$ નું વિકલન શું થાય?

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