## Calculus (3rd Edition)

Assume that $f'(x)=f(x)^2$, then we have $$g'(x)=\frac{1}{f'(g(x))}=\frac{1}{f(g(x))^2}=\frac{1}{x^2}$$ where $g(x)$ is the inverse of $f(x)$.