# 8. Find dy/dx  in the following: $\sin ^2 x + \cos ^ 2 y = 1$

G Gautam harsolia

Given function is
$\sin ^2 x + \cos ^ 2 y = 1$
We can rewrite it as
$\cos ^ 2 y = 1-\sin^2x$
Now, differentiation w.r.t. x is
$\frac{d(\cos^2y)}{dx} = \frac{d(1-\sin^2x)}{dx}$
$2\cos y (-\sin y)\frac{dy}{dx} = -2\sin x \cos x\\ \frac{dy}{dx} = \frac{2\sin x\cos x}{2\sin y \cos y} = \frac{\sin 2x }{\sin 2y} \ \ \ \ \ \ (\because2\sin a \cos a = \sin 2a)$
Therefore, the answer is  $\frac{\sin 2x}{\sin 2y }$

Exams
Articles
Questions