Proof that the product of two differentiable functions is also differentiable Let $f:A\subset \mathbb{R^p}\rightarrow\mathbb{R^q}$ and $\phi:A\subset \mathbb{R^p}\rightarrow\mathbb{R}$ differentiable in $c\in A$. 
I have to prove that $g(x)=\phi (x)f(x)$ is differentiable, where
$Dg(c)u=(D \phi(c)u)f(c)+\phi(c)(Df(c)u)$ for any $u \in \mathbb{R^p}$.
I have done the following:
$g(x)=\phi (x)f(x)$ is differentiable if and only if:
$$\lim_{x\to c}\frac{||g(x)-g(c)-Dg(c)(x-c)||}{||x-c||}=0$$
$$\lim_{x\to c}\frac{||\phi (x)f(x)-\phi (c)f(c)-D \phi(c)(x-c))f(c)-\phi(c)(Df(c)(x-c))||}{||x-c||}=\lim_{x\to c}\frac{||\phi (x)f(x)-\phi (c)f(c)-\phi(c)f(x)+ \phi(c)f(x)-D \phi(c)(x-c))f(c)-\phi(c)(Df(c)(x-c))||}{||x-c||}< \lim_{x\to c}\frac{|\phi(c)| ||f(x)-f(c)-(Df(c)(x-c))||}{||x-c||} +
 \lim_{x\to c}\frac{||\phi (x)f(x)-\phi(c)f(x)-D \phi(c)(x-c))f(c)||}{||x-c||}<|\phi(c)|\lim_{x\to c}\frac{ ||f(x)-f(c)-(Df(c)(x-c))||}{||x-c||} + \lim_{x\to c}\frac{||\phi (x)f(x)-\phi(c)f(x)-D \phi(c)(x-c))f(c)||}{||x-c||}<\lim_{x\to c}\frac{||\phi (x)f(x)-\phi(c)f(x)-D \phi(c)(x-c))f(c)||}{||x-c||}...$$
What can I do with the second part? thank you very much!
 A: You can prove a lemma which says that differentiable implies continuous in your context. Then, the $\phi(x)$ terms naturally factor out in view of the identity $\lim_{x \rightarrow c} f(x) = f(c)$. Of course, you also need to use the differentiability of $\phi$, but I gather you are aware of this as you used the analog for $f$ already in the first half. Moreover, I would encourage you to also check that your proposed derivative is linear. Usually, the definition requires a linear function which satisfies the Frechet quotient. Nice work thus far.
A: Let f(x) and g(x) be two functions differentiable at $x_0$. We want to show that;
$$(fg)'(x)=f(x)'g(x) + f(x)g(x)'$$
$$lim_{x \to x_0}\frac{f(x)g(x)-f(x_0)g(x_0)}{x-x_0}=lim_{x \to x_0}\frac{f(x)g(x)-f(x)g(x_0)+f(x)g(x_0)-f(x_0)g(x_0)}{x-x_0}$$
$$=lim_{x \to x_0}\frac{f(x)(g(x)-g(x_0))+g(x_0)(f(x)-f(x_0))}{x-x_0}$$
=$$lim_{x \to x_0}\frac{f(x)(g(x)-g(x_0))}{x-x_0}+\frac{(g(x_0)(f(x)-f(x_0))}{x-x_0} $$
Now if the function is differentiable at $x_0$, it is also continuous there.
So we end up getting
$$lim_{x \to x_0}f(x)=f(x_0)$$
and the same thing with g(x).  So we'll end up with:
=$$lim_{x \to x_0}\frac{f(x)(g(x)-g(x_0))}{x-x_0}+\frac{(g(x)(f(x)-f(x_0))}{x-x_0} =f(x_0)g'(x_0)+g(x_0)f'(x_0)$$
