Question

x√1-y²+y√1-x² =a, | x | <1 , | y | < 1,Find dy/dx for the given function y wherever defined

Answer 1

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Answer 2

Given, $x\sqrt{1-y^2}+ysqrt{1-x^2}=a$

differentiate with respect to x,

$\frac{d\{x\sqrt{1-y^2}\}}{dx}+\frac{d\{y\sqrt{1-x^2}\}}{dx}=\frac{da}{dx}$

=>$\sqrt{1-y^2}\frac{dx}{dx}+x\frac{d\sqrt{1-y^2}}{dx}+y\frac{d\sqrt{1-x^2}}{dx}+\sqrt{1-x^2}\frac{dy}{dx}=0$

=> $\sqrt{1-y^2}+\frac{-xy}{\sqrt{1-y^2}}\frac{dy}{dx}+\frac{-xy}{\sqrt{1-x^2}}+\sqrt{1-x^2}\frac{dy}{dx}=0$

=> $\frac{\sqrt{(1-x^2)(1-y^2)}-xy}{\sqrt{1-x^2}}+\frac{\sqrt{(1-x^2)(1-y^2)}-xy}{\sqrt{1-y^2}}\frac{dy}{dx}=0$

=> $\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0$

=> $\frac{dy}{dx}=-\sqrt{\frac{1-y^2}{1-x^2}}$