$\frac{d}{dx} \tan^{-1} \left( \frac{1-x}{1+x} \right) = $ . . . . . .

$\frac{d}{dx}\left( \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \right) = $

ધારો કે $f(\theta) = \sin \left(\tan^{-1} \left(\frac{\sin \theta}{\sqrt{\cos 2\theta}} \right) \right)$,જ્યાં $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$ છે. તો $\frac{d}{d(\tan \theta)}(f(\theta))$ નું મૂલ્ય શોધો.

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

$\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$ નું વિકલન શું છે?

$\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] = $