$f,g:\mathbb{R}^n\rightarrow \mathbb{R}^m$ are differentiable. Show that $D_{x_0}(⟨f,g⟩)(h)=⟨D_{x_0} f(h),g(x_0)⟩+⟨f(x_0),D_{x_0}g(h)⟩$, where $⟨·, ·⟩$ is the standard inner product on $\mathbb{R}^m$.
I tried to use the fact that $<f,g>=\frac{\|f+g\|^2-\|f\|^2-\|g\|^2}{2}$, and I have already proved for $f(x)=<x,x>, Df_{x_0}(h)=2<x_0,h>$. So $D_{x_0}(<f,g>)(h)=(D_{x_0}\frac{\|f+g\|^2}{2}-D_{x_0}\frac{\|f\|^2}{2}-D_{x_0}\frac{\|g\|^2}{2})(h)$. I do not know how to do from here. Can anyone help me with it? Thanks.
Consider the two functions. . .
$A : \mathbb R^n \to \mathbb R^{2m}$
$A (h) = (f(h),g(h))$
That is to say the vector with first $m$ entries equal to the entries of $f(h)$ and second $m$ entries equal to the entries of $g(h)$.
$B : \mathbb R^{2m} \to \mathbb R$
$B (x,y) = \langle x,y\rangle$
What you are trying yo differentiate is the function. . .
$B \circ A :\mathbb R^{n} \to \mathbb R$
$h \mapsto B(A(h))$.
There is a rule to compute the derivative of a compound function, called the chain rule:
$D_x(B \circ A) = (D_{A(x)}B)( D_x A)$. . .
Where the product on the RHS is the matix product. Because of the dimensions of the functions involved, the matrix above turns out to be $1\times1$ or in other words a number. You can compute it to equal what you are given on the RHS.