Prove that there is a differentiable function $f: \mathbb{R} \to \mathbb{R}$ satisfying $$[f(x)]^5+f(x)+x = 0$$ for all $x \in \mathbb{R}$. Find $f'(x)$.
Seeing how this is a functional equation, I think we might be able to use induction or some other technique to determine information about $f$. We have that $$f(0)\left([f(0)]^4+1\right) = 0 \implies f(0) = 0$$ How might else we find a way to find $f'$ or to prove that it exists?
Such a function $f$ is simply the inverse of $g(x) = -x^5 - x$, which is monotonic and differentiable with $g'(x) = -5x^4 - 1$ nonzero everywhere. Hence $f$ is differentiable everywhere by the inverse function theorem, and $f'$ is given by the chain rule.