Question

If $y=x{e}^{xy}$ then show that $\frac{dy}{dx}=\frac{y(1+xy)}{x(1-xy)}$

Answer 1

Given $y=x{e}^{xy}$

Applying logarithm on both sides

$⟹\log y=\log x+xy$

Differentiating on both sides

$⟹\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+y+x\frac{dy}{dx}$

$⟹(\frac{1}{y}-x)\frac{dy}{dx}=\frac{1}{x}+y$

$⟹(\frac{1-xy}{y})\frac{dy}{dx}=\frac{1+xy}{x}$

$⟹\frac{dy}{dx}=\frac{y(1+xy)}{x(1-xy)}$