Question

If $x^3-y^3+\tan \left(xy\right)=\log (x+y)$, find $\frac{dy}{dx}$

Answer 1

$x^3-y^3+\tan \left(xy\right)=\log (x+y)$

Differentiating w.r.t. $x$

$3x^2-3y^2\frac{dy}{dx}$ $+{\sec }^2\left(xy\right)(y+x\frac{dy}{dx})$ $=\frac{1}{x+y}$ $(1+\frac{dy}{dx})$

$\Rightarrow 3x^2+y{\sec }^2\left(xy\right)-\frac{1}{x+y}$ $=(3y^2-x{\sec }^2(xy)+\frac{1}{x+y})\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{3x^2+y{\sec }^2\left(xy\right)-\frac{1}{x+y}}{3y^2-x{\sec }^2\left(xy\right)+\frac{1}{x+y}}$