જો y = tan^{-1}left( frac{x}{1 + sqrt{1 - x^2 | vedclass

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

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

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(A) આપેલ છે કે $y = 	an^{-1}\left( \frac{x}{1 + \sqrt{1 - x^2}} ight)$.$x = \sin 	heta$ આદેશ લેતા,જ્યાં $	heta = \sin^{-1} x$.તેથી $y = 	an^{-1}\left( \frac{\sin 	heta}{1 + \sqrt{1 - \sin^2 	heta}} ight) = 	an^{-1}\left( \frac{\sin 	heta}{1 + \cos 	heta} ight)$.ત્રિકોણમિતીય નિત્યસમ $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ અને $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$ નો ઉપયોગ કરતા:$y = 	an^{-1}\left( \frac{2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \cos^2 \frac{	heta}{2}} ight) = 	an^{-1}\left( 	an \frac{	heta}{2} ight) = \frac{	heta}{2}$.$	heta = \sin^{-1} x$ મૂકતા,આપણને $y = \frac{1}{2} \sin^{-1} x$ મળે.$x$ ની સાપેક્ષમાં વિકલન કરતા:$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{1}{2\sqrt{1 - x^2}}$.

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

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