Prove a function is differentiable with chain rule Let $A$ be an open set of $\mathbb{R}^n$ and let $f : A \rightarrow \mathbb{R}^m$. Fix $u \in \mathbb{R}^m$ and define $g : A \rightarrow \mathbb{R}$ by $g(x) = f(x) \bullet u$ for all $x \in A$, where $\bullet$ is the inner product. Show that if $f$ is differentiable on $A$ then $g$ is differentiable on $A$ with derivative $g'(a) = f'(a) \bullet u$. 
I am trying to prove this with the chain rule. If $h : \mathbb{R}^m \rightarrow \mathbb{R}$ is defined by $h(x) = x \bullet u$ then $h$ is differentiable, so $h \circ f = g$ is differentiable. I am having a hard time showing that $g'(a) = f'(a) \bullet u$. 
 A: Just going to use the definition of the total derivative here.
Assume $x \in A$. Because $f$ is differentiable, we can say that:
$$\lim_{t \to 0} \frac{||f(x + t) - f(x) -f'(x)t||}{||t||} = 0$$
Using the definition of the derivative, we can say that $g$ is differentiable on $A$ if: 
$$\lim_{t \to 0} \frac{||g(x + t) - g(x) -g'(x)t||}{||t||} = 0$$
For some linear transformation: $g'(x): A \to R$.
Put $g'(x) = f'(x)\cdot u$
$$0 \leq \lim_{t \to 0} \frac{||g(x + t) - g(x) -g'(x)t||}{||t||} = \lim_{t \to 0} \frac{||f(x + t)\cdot u - f(x)\cdot u -(f'(x) \cdot u)t||}{||t||}$$
$$\lim_{t \to 0} \frac{||f(x + t)\cdot u - f(x)\cdot u -(f'(x) \cdot u)t||}{||t||} = \lim_{t \to 0} \frac{||(f(x + t) - f(x) -f'(x)t)\cdot u||}{||t||}$$
$$\lim_{t \to 0} \frac{||(f(x + t) - f(x) -f'(x)t)\cdot u||}{||t||} \leq \lim_{t \to 0} \frac{||f(x + t) - f(x) -f'(x)t||}{||t||}||u||$$
By Cauchy Schwartz.
$$\lim_{t \to 0} \frac{||f(x + t) - f(x) -f'(x)t||}{||t||}||u|| = 0 * ||u|| = 0$$
Therefore, $g(x)$ is differentiable on $A$ and $g'(x) = f'(x) \cdot u$
A: Your formula $g'(a) = f'(a) \bullet u$ is a little bit off, since $f'(a)$ is a linear map, and $u$ is a vector.
Since $h(y):=u\cdot y$ is linear we have
$$\eqalign{g(a+X)-g(a)&=u\cdot\bigl(f(a+X)\bigr)-u\cdot\bigl(f(a)\bigr)=u\cdot\bigl(f(a+X)-f(a)\bigr)\cr &=u\cdot\bigl(df(a).X+o(|X|)\bigr)=u\cdot\bigl(df(a).X\bigr)+o(|X|)\cr &=\bigl(df(a)^{\top}u\bigr)\cdot X+o(|X|)\qquad(X\to0) .\cr} $$
This shows that the gradient of the scalar function $g$ is given by
$$\nabla g(a)=df(a)^{\top}\>u\ \ .$$
