Question

Find $\frac{dy}{dx}$, if $2x^3+2y^3+x^2+y^2+4=\sin x$

Answer 1

$2x^3+2y^3+x^2+y^2+4=\sin x$
The given function is an implicit function. Therefore, differentiating with respect to $x$ gives:
$6x^2+6y^2\frac{dy}{dx}+2x+2y\frac{dy}{dx}+0=\cos x$
$\Rightarrow (6y^2+2y)\frac{dy}{dx}=\cos x-6x^2-2x$
$\therefore \frac{dy}{dx}=\frac{\cos x-6x^2-2x}{6y^2+2y}$