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

જો $y = \tan^{-1} \left( \frac{\ln(e/x^2)}{\ln(ex^2)} \right) + \tan^{-1} \left( \frac{3 + 2 \ln x}{1 - 6 \ln x} \right)$ હોય,તો $\frac{d^2y}{dx^2} =$

જો $y = \operatorname{Sin}^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$ અને $\frac{-3\pi}{2} < x < \frac{-\pi}{2}$ હોય,તો $\frac{dy}{dx} = $

$\sec ^2(\tan ^{-1} 3) + \operatorname{cosec}^2(\cot ^{-1} 3) = $ . . . . . . .

$x$ ની કિંમત શોધો, જ્યાં $x>0$ અને $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$ છે.

$\sin ^{-1}\left(\frac{12}{13}\right)+\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right)=$