Showing differentiability of product of two functions Let $\Omega$ be a nonempty open subset of $\mathbb{R}^n$ and let $f:\Omega\to \mathbb R$ be a differentiable function at $\mathbf{x}_0\in \Omega$ such that $f(\mathbf{x}_0)=0$. If $g$ is continuous at $x_0$, then I have to show that $fg$ on $\Omega$ defined by $fg(\mathbf{x})=f(\mathbf{x})g(\mathbf{x})$ is differentiable at $\mathbf{x}_0$. Any help is appreciated.
 A: If $g$ is also differentiable at $x_0$, you can do it by using only the definition of differentiability.
$f$ (resp. $g$) is differentiable at $x_0 \in \Omega$ if there exist a linear map $\mathrm{D}_{x_0}f$ (resp. $\mathrm{D}_{x_0}g$) from $\mathbb{R}^n$ to $\mathbb{R}$ such that, for any $h \in \mathbb{R}^n$ (sufficiently small) :
$$ f(x_0+h) = f(x_0) + \mathrm{D}_{x_0}f \cdot h + o\big( \Vert h \Vert \big) $$
resp. $g(x_0+h) = g(x_0) + \mathrm{D}_{x_0}g \cdot h + o\big( \Vert h \Vert \big)$.
Instead of writing $o\big( \Vert h \Vert \big)$, write $\Vert h \Vert \varepsilon(\Vert h \Vert)$ where $\varepsilon$ is a function which goes to $0$ as $\Vert h \Vert \to 0$. Then, $f(x_0+h)g(x_0)$ writes :
$$ \small{ f(x_0+h)g(x_0+h) = g(x_0)\mathrm{D}_{x_0}g \cdot h + \underbrace{\big[ \mathrm{D}_{x_0}f \cdot h \times \mathrm{D}_{x_0}g \cdot h + g(x_0)\Vert h \Vert \varepsilon(\Vert h \Vert) + \big( \Vert h \Vert \varepsilon(\Vert h \Vert) \big)^2 \big]}_{=(\star)} } $$ 
It remains to prove that the term between bracket is $o\big( \Vert h \Vert \big)$. For that, use triangular inequality an note that, since $\mathrm{D}_{x_0}f$ and $\mathrm{D}_{x_0}g$ are linear, we have :
$$ \vert \mathrm{D}_{x_0}f \cdot h \vert \leq \Vert \mathrm{D}_{x_0}f \Vert \Vert h \Vert \quad \text{and} \quad \vert \mathrm{D}_{x_0}g \cdot h \vert \leq \Vert \mathrm{D}_{x_0}g \Vert \Vert h \Vert. $$
It follows that $fg$ is differentiable at $x_0$ and, for all $h$ : 
$$ \mathrm{D}_{x_0}(fg) \cdot h = g(x_0) \mathrm{D}_{x_0}f \cdot h. $$
A: $fg$ is differentiable at $x_0$ if $\lim_{x\rightarrow x_0} \frac{fg(x)-fg(x_0)}{d(x, x_0)}$ exists.
$fg(x_0)=0$ because $f(x_0)$ is $0$, so the above limit is
$\lim_{x\rightarrow x_0} \frac{fg(x)}{d(x, x_0} = \lim_{x\rightarrow x_0} \left(\frac{f(x)}{d(x, x_0)} \cdot g(x)\right)$
Because $f$ is differentiable and $f(x_0)=0$, we have that $\lim_{x\rightarrow x_0}\frac{f(x)}{d(x, x_0)} = \lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{d(x, x_0)}$ exists, and because $g$ is continuous, $\lim_{x\rightarrow x_0}g(x)$ exists, and so the limit of the product of these functions exists as well. 
Hope this helps!
EDIT: 
Taking the decomposition of $f$ as $f(x) = L(x) + \alpha(x)$, where $\alpha(x)$ goes to 0 as $x \rightarrow x_0$ (to say that $f$ has a linear approximation near $x_0$) and $L$ is linear, then we have that $fg(x) = (L(x) + \alpha(x))g(x)=L(x)g(x) + g(x)\alpha(x)$. 
Take $H(x) = L(x)g(x_0)$. Since $L(x_0) = f(x_0) =0$ and $g$ is bounded, we can show that $L(x)g(x)$ gets $\varepsilon$-close to $H(x)$ around $x_0$. 
Taking $\beta(x) = \alpha(x)g(x)$, and using that $g$ is bounded, we have that $\beta (x) \rightarrow 0$. These facts taken together give that $H(x) + \beta (x)$ is the desired approximation of $fg$ as a linear function.
