Question

Find $\frac{dy}{dx}$
$\tan (x+y)=x^2+y^2$

Answer 1

$\tan (x+y)=x^2+y^2$

Diff. wrt. $\left(x\right)$, we get

$\frac{d}{dx}\left(\tan (x+y)\right)=\frac{d}{dx}(x^2+y^2)$

$\therefore {\sec }^2(x+y)[1+\frac{dy}{dx}]=2x+2y\frac{dy}{dx}$

$\therefore {\sec }^2(x+y)+{\sec }^2(x+y)\frac{dy}{dx}=2x+2y\frac{dy}{dx}$

$\therefore {\sec }^2(x+y)-2x=\frac{dy}{dx}(2y-{\sec }^{2}x+y)$

$\therefore \frac{dy}{dx}=\frac{{\sec }^2(x+y)-2x}{2y-{\sec }^2(x+y)}$

$\therefore \frac{dy}{dx}=\frac{{\sec }^2(x+y)-2x}{2y-{\sec }^2(x+y)}$