Q:

Chain rule and multivariable derivatives

Given $F\colon\mathbb{R}^2\to\mathbb{R}$ defined as $F(x,y) = x^2 y + y^3 + 2x-1$ and $g\colon\mathbb{R}\to\mathbb{R}$ such that $g(0) = 1$ and $F(x, g(x)) = x$, find $g'(0)$.

A:

$(x^2)g(x) + (g(x)^3) + 2x -1 = x$
Using implicit derivation
$2xg(x) + x^2g'(x) + 3g(x)^2g'(x) +2 = 1$
Can you finish it?

