Let $f: U \longrightarrow \mathbb{R}$ be a function defined in open $U \subset \mathbb{R}^{m}$. Define $f^{k}(x) = f(x)^{k}$. Prove that $f^{k}$ is differentiable and $df^{k}(x)v = kf^{k-1}(x)df(x)v$ for $x \in U$ and $v \in \mathbb{R}^{m}$.
I know that $f$ is differentiable in $a$: $$df(a)v = \frac{\partial f}{\partial v}$$ and $$f(a+v)=f(a) + \frac{\partial f}{\partial x_{1}}(a)\alpha_{1} + ... + \frac{\partial f}{\partial x_{n}}(a)\alpha_{n} + r(v)$$ with $\displaystyle\lim_{v\to 0} \frac{r(v)}{|v|} = 0$.
I've tried writing $$f^{k}(a+v)=f^{k}(a) + kf^{k-1}(a)df(a)x_{1}\alpha_{1} + ... + kf^{k-1}(a)df(a)x_{n}\alpha_{n} + r(v)$$ and show that $\displaystyle \lim_{v \to 0} \frac{r(v)}{|v|}=0$. But, I don't know if $f$ is differentiable in $a$.
Any hints? Thanks for the advance!r
Chain Rule, no?
For $\varphi(u)=u^{k}$, we are dealing with the differential of $\varphi\circ f(x)$ for $x\in U$. Since $D\varphi(f(x))(h)=k(f(x))^{k-1}(h)$ exists, then so is $D(\varphi\circ f)(x)$ by Chain Rule, and \begin{align*} D(\varphi\circ f)(x)(h)=D\varphi(f(x))Df(x)(h)=k(f(x))^{k-1}(f'(x)h). \end{align*}