Question

Find $\frac{dy}{dx}$ : $4x^2-\log x=y^2+\sin x-\cos x$

Answer 1

$4x^2-\log x=y^2+\sin x-\cos x$

$8x-\frac{1}{x}=2y\frac{dy}{dx}+\cos x+\sin x$

$2y\frac{dy}{dx}=8x-\frac{1}{x}-\sin x-\cos x$

$\frac{dy}{dx}=\frac{1}{2y}(8x-\frac{1}{x}-\cos x-\sin x)$