Theorem: If $f$ and $g$ are differentiable then $\frac{d}{dx}(f(x)g(x))=\frac{d}{dx}(f(x))g(x)+\frac{d}{dx}(g(x))f(x)$
Proof: We apply the definition of $\frac{d}{dx}$, but then we need a trick, and we also use the result that if a function is differentiable then it is continuous.

Let $F(x)=f(x)g(x)$. Then

$\frac{d}{dx}(F(x))=\lim_{h\to 0}\ \frac{1}{h}(F(x+h)\ -\ F(x))$

$=\lim_{h\to 0}\ \frac{1}{h}(f(x+h)g(x+h)\ -\ f(x)g(x))$

$=\lim_{h\to 0}\ \frac{1}{h}(f(x+h)g(x+h)\ -\ f(x)g(x+h)+f(x)g(x+h)\ -\ f(x)g(x))$ (the trick!)

$=\lim_{h\to 0}(\frac{1}{h}(f(x+h)\ -\ f(x))g(x+h)+(g(x+h)\ -\ g(x))f(x))$

$=\lim_{h\to 0}(\frac{1}{h}(f(x+h)\ -\ f(x))g(x+h)+\frac{1}{h}(g(x+h)\ -\ g(x))f(x))$

$=\lim_{h\to 0}\ \frac{1}{h}(f(x+h)\ -\ f(x))g(x+h)+\lim_{h\to 0}\ \frac{1}{h}(g(x+h)\ -\ g(x))f(x)$

$=\lim_{h\to 0}\ \frac{1}{h}(f(x+h)\ -\ f(x))\lim_{h\to 0}\ g(x+h)+\lim_{h\to 0}\ \frac{1}{h}(g(x+h)\ -\ g(x))\lim_{h\to 0}\ f(x)$


In the last step we used the fact that $\lim_{h\to 0}\ g(x+h)=g(\lim_{h\to 0}(x+h))=g(x)$ since $g$ is differentiable and hence continuous. Also $\lim_{h\to 0}\ f(x)=f(x)$ since $f(x)$ is constant with respect to $h$.