begin{aligned} & text{જો } y = tan^{-1} left{ | vedclass

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(A) આપેલ છે કે,$y = 	an^{-1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} ight\} + \sin \left\{ 2 	an^{-1} \sqrt{\frac{1 - x}{1 + x}} ight\}$.ધારો કે $x

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$\frac{1 - 2x}{2 \sqrt{1 - x^2}}$

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(A) આપેલ છે કે,$y = 	an^{-1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} ight\} + \sin \left\{ 2 	an^{-1} \sqrt{\frac{1 - x}{1 + x}} ight\}$.ધારો કે $x = \cos 2	heta$,તેથી $	heta = \frac{1}{2} \cos^{-1} x$.$x = \cos 2	heta$ ને પદાવલિમાં મૂકતા:$y = 	an^{-1} \left\{ \frac{\cos 2	heta}{1 + \sin 2	heta} ight\} + \sin \left\{ 2 	an^{-1} \sqrt{\frac{1 - \cos 2	heta}{1 + \cos 2	heta}} ight\}$$y = 	an^{-1} \left\{ \frac{\cos^2 	heta - \sin^2 	heta}{(\cos 	heta + \sin 	heta)^2} ight\} + \sin \left\{ 2 	an^{-1} (	an 	heta) ight\}$$y = 	an^{-1} \left\{ \frac{\cos 	heta - \sin 	heta}{\cos 	heta + \sin 	heta} ight\} + \sin 2	heta$$y = 	an^{-1} \left\{ 	an \left( \frac{\pi}{4} - 	heta ight) ight\} + \sin 2	heta$$y = \frac{\pi}{4} - 	heta + \sin 2	heta$$	heta = \frac{1}{2} \cos^{-1} x$ અને $\sin 2	heta = \sqrt{1 - x^2}$ મૂકતા:$y = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x + \sqrt{1 - x^2}$$x$ ની સાપેક્ષ વિકલન કરતા:$\frac{dy}{dx} = 0 - \frac{1}{2} \left( -\frac{1}{\sqrt{1 - x^2}} ight) + \frac{1}{2\sqrt{1 - x^2}} (-2x)$$\frac{dy}{dx} = \frac{1}{2\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}} = \frac{1 - 2x}{2\sqrt{1 - x^2}}$.

$\begin{aligned} & \text{જો } y = \tan^{-1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\} \\ & + \sin \left\{ 2 \tan^{-1} \sqrt{\frac{1 - x}{1 + x}} \right\} \text{ હોય, તો } \frac{dy}{dx} = \end{aligned}$

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