Question

Find $\frac{dy}{dx}$ :

$x+y^2=\log y+x^2$

Answer 1

$x+y^2=\log y+x^2$

Differentiating w.r.t. x, we get,

$1+2y\frac{dy}{dx}=\frac{1}{y}.\frac{dy}{dx}+2x$

$\frac{dy}{dx}(2y-\frac{1}{y})=2x-1$

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

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