{tan ^{ - 1}}left( frac{{2x}}{{1 - {x^2}}} ri | vedclass

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(A) ધારો કે ${y_1} = {	an ^{ - 1}}\left( \frac{{2x}}{{1 - {x^2}}} ight)$ અને ${y_2} = {\sin ^{ - 1}}\left( \frac{{2x}}{{1 + {x^2}}} ight)$.$x = \

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(A) ધારો કે ${y_1} = {	an ^{ - 1}}\left( \frac{{2x}}{{1 - {x^2}}} ight)$ અને ${y_2} = {\sin ^{ - 1}}\left( \frac{{2x}}{{1 + {x^2}}} ight)$.$x = 	an 	heta$ આદેશ લેતા,જ્યાં $	heta = {	an ^{ - 1}}x$.તેથી,${y_1} = {	an ^{ - 1}}\left( \frac{{2	an 	heta }}{{1 - {{	an }^2}	heta }} ight) = {	an ^{ - 1}}(	an 2	heta) = 2	heta = 2{	an ^{ - 1}}x$.તે જ રીતે,${y_2} = {\sin ^{ - 1}}\left( \frac{{2	an 	heta }}{{1 + {{	an }^2}	heta }} ight) = {\sin ^{ - 1}}(\sin 2	heta) = 2	heta = 2{	an ^{ - 1}}x$.હવે,આપણે વિકલન $\frac{{d{y_1}}}{{d{y_2}}}$ શોધવાનું છે.કારણ કે ${y_1} = 2{	an ^{ - 1}}x$ અને ${y_2} = 2{	an ^{ - 1}}x$,તેથી ${y_1} = {y_2}$ થાય.તેથી,$\frac{{d{y_1}}}{{d{y_2}}} = \frac{d}{{d{y_2}}}({y_2}) = 1$.

${\tan ^{ - 1}}\left( \frac{{2x}}{{1 - {x^2}}} \right)$ નું ${\sin ^{ - 1}}\left( \frac{{2x}}{{1 + {x^2}}} \right)$ ની સાપેક્ષ વિકલન સહગુણક શોધો.

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