> [!NOTE] Theorem (The product rule for derivative) > Suppose $f,g:I \to \mathbb{R}$ are defined on the open interval $I$ and are [[Fréchet Differentiation|differentiable]] at $c\in I$. Then $fg$ is differentiable at $c$ and its [[Derivative of Real Function|derivative]] is given by $(fg)'(c)=f'(c)g(c)+f(c)g'(c)$ **Proof**: We have, $\begin{align} \frac{f(x)g(x) - f(c)g(c)}{x-c} &= \frac{(f(x)-f(c))g(x) + f(c)(g(x)-g(c))}{x-c} \\ & = \frac{f(x)-f(c)}{x-c} g(x) + f(c) \frac{g(x)-g(c)}{x-c} \end{align}$By [[Differentiablity implies Continuity]] and [[Algebra of Limits of Real Functions|algebra of limits]], as $x\to c$ this expression approaches $f'(c)g(c)+f(c)g'(c).$