Let's say that $g$ is differentiable in $x_0$ and $f$ is differentiable in $y_0 = g(x_0)$ We define: $$ \varphi(y) = \left \{ \begin{array}{ll} \frac{f(y) - f(y_0)}{y-y_0} & \textrm{if $y \neq y_0$} \\ f'(y_0) & \textrm{if $y = y_0$} \end{array} \right. $$

Since $f$ is differentiable in $y_0$ $\varphi$ is continuous. We observe that, for $x \neq x_0$ $$ \frac{f(g(x))-f(g(x_0))}{x-x_0} = \varphi(g(x)) \frac{g(x) - g(x_0)}{x-x_0}, $$ in fact, if $g(x) \neq g(x_0)$ it follows at once from the definition of $\varphi$ while if $g(x) = g(x_0)$ both members of the equation are 0.

Since $g$ is continuous in $x_0$ and $\varphi$ is continuous in $y_0$ $$ \lim_{x \to x_0} \varphi(g(x)) = \varphi(g(x_0)) = f'(g(x_0)), $$ hence \begin{eqnarray*} (f \circ g)'(x_0) &=& \lim_{x\to x_0} \frac{f(g(x))-f(g(x_0))}{x-x_0} \\ &=& \lim_{x\to x_0} \varphi(g(x)) \frac{g(x) - g(x_0)}{x-x_0} \\ &=& f'(g(x_0))g'(x_0). \end{eqnarray*}