proof of product rule ProofOfProductRule 2002-02-24 20:53:12 2004-10-15 17:15:15 Proof product rule 0 \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \newcommand{\D}[1]{\ensuremath{\mathrm{d}#1}} We begin with two differentiable functions $f(x)$ and $g(x)$ and show that their product is differentiable, and that the derivative of the product has the desired form. By simply calculating, we have for all values of $x$ in the domain of $f$ and $g$ that \begin{eqnarray*} \frac{\D{}}{\D{x}}\left[f(x)g(x)\right] & = & \lim_{h\to0}\frac{f(x+h)g(x+h) - f(x)g(x)}{h} \\ & = & \lim_{h\to0}\frac{f(x+h)g(x+h) + f(x+h)g(x) - f(x+h)g(x) - f(x)g(x)}{h} \\ & = & \lim_{h\to0}\left[f(x+h)\frac{g(x+h)-g(x)}{h} + g(x)\frac{f(x+h)-f(x)}{h}\right] \\ & = & \lim_{h\to0}\left[f(x+h)\frac{g(x+h)-g(x)}{h}\right] + \lim_{h\to0}\left[g(x)\frac{f(x+h)-f(x)}{h}\right] \\ & = & f(x)g'(x) + f'(x)g(x). \end{eqnarray*} The key argument here is the next to last line, where we have used the fact that both $f$ and $g$ are differentiable, hence the limit can be distributed across the sum to give the desired equality.