Prove that $F(x)=\alpha(x)f(x)$ is differentiable and compute the derivative For an assignment, I have to solve this problem, but I just can't figure out how to continue. I already figured it out for the case $F(x)=f(x)+g(x)$ but this one I just cannot figure out. Would be great if anyone could let me know what I'm missing!
Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ and $\alpha:\mathbb{R}^n\rightarrow\mathbb{R}$ be differentiable functions. Prove that the following function is differentiable and compute the derivative:
$$F:\mathbb{R}^n\rightarrow\mathbb{R}^m$$
$$F(x)=\alpha(x)f(x)$$
We were given that the definition of a differentiable function is that there exists a linear transformation $A:\mathbb{R}^n\rightarrow\mathbb{R}^m$ such that 
$$\lim_{h\rightarrow 0}\frac{\lVert f(x+h)-f(x)-Ah \rVert}{\lVert h \rVert}=0$$
So I tried to use this definition for $\alpha(x)$ and $f(x)$: Since $\alpha(x)$ is differentiable there exists an $A_1$ such that 
$$\lim_{h\rightarrow 0}\frac{\lVert \alpha(x+h)-\alpha(x)-A_1h \rVert}{\lVert h \rVert}=0$$
And since $f(x)$ is differentiable, there exists an $A_2$ such that
$$\lim_{h\rightarrow 0}\frac{\lVert f(x+h)-f(x)-A_2h \rVert}{\lVert h \rVert}=0$$
Then by multiplying these it should follow that there exists some $A_3$ which is a combination of $A_1$ and $A_2$ that shows that $F(x)$ is differentiable. But when I multiply these, it just becomes one big mess:
$$\lim_{h\rightarrow 0}\frac{\lVert \alpha(x+h)f(x+h)-\alpha(x+h)f(x)-\alpha(x+h)A_2h-\alpha(x)f(x+h)+\alpha(x)f(x)+\alpha(x)A_2h-A_1hf(x+h)+A_1hf(x)+A_1A_2h^2 \rVert}{\lVert h \rVert^2}$$
Of course you can now say that $F(x+h)=\alpha(x+h)f(x+h)$ and $F(x)=\alpha(x)f(x)$ and substitute those, but that doesn't get me that far.
Any help on further steps is much appreciated!
 A: Multiplying them is a big mess, and there is no reason to do it. Recall from single variable calculus that to find the derivative of a product, we do not simply multiply the derivatives of each factor. Instead, we have the product rule. In order prove that the derivative exists, it helps to have a guess for what the derivative is. To do this, we are going to invent a product rule for this type of function. The linear operator we will guess is $$A_3(x)=f(x)\cdot A_1(x)+\alpha(x)\cdot A_2(x).$$ (observe that I have replaced $A_i$ with $A_i(x)$, since these linear operators will depend on $x$). Now let us prove that our product rule is right.
\begin{align*}
&\lim_{h\to 0}\frac{\Vert F(x+h)-F(x)-A_3(x)h\Vert}{\Vert h\Vert}\\
&=\lim_{h\to 0}\frac{\Vert \alpha(x+h)f(x+h)-\alpha(x)f(x)-f(x)A_1(x)h-\alpha(x)A_2(x)h\Vert}{\Vert h\Vert}\\
&=\lim_{h\to 0}\frac{\Vert \alpha(x+h)f(x+h)-\alpha(x+h)f(x)+\alpha(x+h)f(x)-\alpha(x)f(x)-f(x)A_1(x)h-\alpha(x)A_2(x)h\Vert}{\Vert h\Vert}\\
&\leq\lim_{h\to 0}\frac{\Vert \alpha(x+h)f(x+h)-\alpha(x+h)f(x)-\alpha(x)A_2(x)h\Vert}{\Vert h\Vert}+\\
&\qquad\lim_{h\to 0}\frac{\Vert\alpha(x+h)f(x)-\alpha(x)f(x)-f(x)A_1(x)h\Vert}{\Vert h\Vert}\\
\end{align*}
Now show why both of these limits are zero.
A: Are you familiar with the little $o$ notation? Write $f(x+h)=f(x)+A_1h+o(\Vert h\Vert)$ and $\alpha(x+h)=\alpha(x)+A_2 h +o(\Vert h\Vert)$. Now
$$ F(x+h)=f(x)\alpha(x)+f(x)A_2h+f(x)o(\Vert h\Vert)+A_1h\alpha(x)+A_1ho(\Vert h\Vert)+o(\Vert h\Vert)\alpha(x)+o(\Vert h\Vert)A_2h+o(\Vert h\Vert^2) $$
Taking 
$$\lim_{h \to 0} \frac{\Vert F(x+h)-F(x)-A_3h \Vert }{\Vert h\Vert}  $$
you can guess what $A_3$ has to be: $A_3=f(x)A_2+\alpha(x)A_1 $.
Let me know if you need further explanation.
