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$f,g:\mathbb{R}^n\rightarrow \mathbb{R}^m$ 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 $=\frac{\|f+g\|^2-\|f\|^2-\|g\|^2}{2}$, and I have already proved for $f(x)=, Df_{x_0}(h)=2$. So $D_{x_0}()(h)=(D_{x_0}\frac{\|f+g\|^2}{2}-D_{x_0}\frac{\|f\|^2}{2}-D_{x_0}\frac{\|g\|^2}{2})(h)$. I do not know how to do from here. Can anyone help me with it? Thanks.

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    It's better to keep the inner product and use its bilinearity. $$\langle f,g\rangle(x_0+h) = \langle f(x_0+h), g(x_0+h)\rangle = \langle f(x_0) + D_{x_0}f(h) + o(h), g(x_0) + D_{x_0}g(h) + o(h)\rangle = \dotsc$$ – 2017-01-31
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    @DanielFischer I tried this, but this does not seem to work – 2017-01-31
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    It does work. What problem did you encounter when you tried that expansion? – 2017-01-31
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    @DanielFischer Should it be $D_{x_0} \langle f,g\rangle (x_0+h)$? – 2017-01-31
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    No. If we distinguished between the affine space $\mathbb{R}^n$ and the vector space, writing $D_{x_0}\langle f,g\rangle (x_0 + h)$ wouldn't even be well-formed. The differential $D_{x_0}\langle f,g\rangle$ takes a vector as argument, and $x_0 + h$ is a point and not a vector if we look at an affine space. If we don't distinguish between affine and vector space, $D_{x_0}\langle f,g\rangle (x_0 + h)$ is well-formed, but it is rarely something one is actually interested in. – 2017-01-31
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    ok, now I get $\lim_{h\to 0} (x_0+h) - (x_0) = ⟨D_{x_0} f(h),g(x_0)⟩+⟨f(x_0),D_{x_0}g(h)⟩+$. How should I continue? – 2017-01-31

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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.