Question

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

Answer 1

​$\mathrm{We}\mathrm{have},\sin \left(xy\right)+\frac{y}{x}=x^2-y^2$
Differentiating with respect to x, we get,
$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(\sin xy\right)+\frac{d}{dx}\left(\frac{y}{x}\right)=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(y^2\right) \\ \Rightarrow \cos \left(xy\right)\frac{d}{dx}\left(xy\right)+\left\{\frac{x{\displaystyle \frac{dy}{dx}}-y{\displaystyle \frac{d}{dx}}\left(x\right)}{x^2}\right\}=2x-2y\frac{dy}{dx} \\ \Rightarrow \cos \left(xy\right)\left\{x\frac{dy}{dx}+y\frac{d}{dx}\left(x\right)\right\}+\left\{\frac{x{\displaystyle \frac{dy}{dx}}-y\left(1\right)}{x^2}\right\}=2x-2y\frac{dy}{dx} \\ \Rightarrow \cos \left(xy\right)\left\{x\frac{dy}{dx}+y\left(1\right)\right\}+\frac{1}{x^2}\left(x\frac{dy}{dx}-y\right)=2x-2y\frac{dy}{dx} \\ \Rightarrow x\cos \left(xy\right)\frac{dy}{dx}+y\cos \left(xy\right)+\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}=2x-2y\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}\left\{x\cos \left(xy\right)+\frac{1}{x}+2y\right\}=\frac{y}{x^2}-y\cos \left(xy\right)+2x \\ \Rightarrow \frac{dy}{dx}\left\{\frac{x^2\cos \left(xy\right)+1+2xy}{x}\right\}=\frac{1}{x^2}\left(y-x^{2}y\cos \left(xy\right)+2x^3\right) \\ \Rightarrow \frac{dy}{dx}=\frac{2x^3+y-x^{2}y\cos \left(xy\right)}{x\left(x^2\cos \left(xy\right)+1+2xy\right)} \end{gathered}$$