Question

Find $\frac{dy}{dx}$ when 𝑥 and 𝑦 are connected by the relation $sin\left(xy\right)+\frac{x}{y}=x^2-y$

Answer 1

Given: $sin\left(xy\right)+\frac{x}{y}=x^2-y$
Differentiating both sides w.r.t. 𝑥
We get,
$\frac{d}{dx}$$\left(sin\right(xy\left)\right)+\frac{d}{dx}$ $\left(\frac{x}{y}\right)=$$\frac{d\left(x^2\right)}{dx}$ $-\frac{dy}{dx}$
$\Rightarrow cosxy\cdot \frac{dxy}{dx}$$+\frac{y.\frac{dx}{dx}-x.\frac{dx}{dy}}{y^2}=2x-\frac{dy}{dx}$

$[\frac{d\left(uv\right)}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}]$

$\Rightarrow \cos \left(xy\right)\cdot [y+x\frac{dy}{dx}]+\frac{y-x\frac{dy}{dx}}{y^2}=2x-\frac{dy}{dx}$
$\Rightarrow \cos \left(xy\right)\cdot [y^3+x{y}^2\frac{dy}{dx}]+y-x\frac{dy}{dx}=2x{y}^2-y^2\frac{dy}{dx}[\because \text{multiplying by}{y}^{2}onbothsides]$
$\Rightarrow \frac{dy}{dx}\left[\cos \right(xy)\cdot x{y}^2-x+y^2]$
$\Rightarrow \frac{dy}{dx}=\left[\frac{2x{y}^2-y-y^3\cos \left(xy\right)}{x{y}^2\cos \left(xy\right)-x+y^2}\right]$