Question

If, $\sin \left(xy\right)+\frac{y}{x}=x^2-y^2$.
Find $\frac{dy}{dx}$.

Answer 1

$\sin xy+\frac{y}{x}=x^2-y^2$

$\Rightarrow \cos xy(x\frac{dy}{dx}+y)+\frac{x\frac{dy}{dx}-y}{x^2}=2x-2y\frac{dy}{dx}$

$\Rightarrow x\cos xy\frac{dy}{dx}+y\cos xy+\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}=2x-2y\frac{dy}{dx}$

$\Rightarrow (x\cos xy+\frac{1}{x}+2y)\frac{dy}{dx}=2x-y\cos xy+\frac{y}{x^2}$

$\Rightarrow \frac{dy}{dx}=\frac{2x-y\cos xy+\frac{y}{x^2}}{(x\cos xy+\frac{1}{x}+2y)}$

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