Product rule
lastmod 2018-08-17created 2015-12-303 edits
The product rule is the rule for differentiating the product of two functions, also called the Leibniz rule.
Given two functions \(f\) and \(g\), the derivative of their product \(fg\)
\[ (fg)'=f'g+fg' \]
is as above.
Proof
Define the function f as \(f(x) = g(x)h(x)\). Finding \(f'(x)\) by the definition of the derivative,
\[ f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \]
\[ = \lim_{\Delta x \to 0} \frac{g(x + \Delta x)h(x + \Delta x) - g(x)h(x)}{\Delta x} \]
\[ = \lim_{\Delta x \to 0} \frac{g(x)h(x + \Delta x) - g(x)h(x) + g(x + \Delta x)h(x + \Delta x) - g(x)h(x + \Delta x)}{\Delta x} \]
\[ = \lim_{\Delta x \to 0} \frac{g(x)(h(x + \Delta x) - h(x)) + h(x + \Delta x)(g(x + \Delta x) - g(x))}{\Delta x} \]
\[ = \lim_{\Delta x \to 0} \left(g(x)\frac{h(x + \Delta x) - h(x)}{\Delta x} + h(x + \Delta x) \frac{g(x + \Delta x) - g(x)}{\Delta x}\right) \]
Here, since \(h(x)\) is continuous in \(x\), the following holds.
\[ \lim_{\Delta x \to 0} h(x + \Delta x) = h(x) \]
Therefore we get the following result.
\[ f'(x) = \lim_{\Delta x \to 0} \left[g(x)\left(\frac{h(x + \Delta x) - h(x)}{\Delta x}\right) + h(x + \Delta x)\left(\frac{g(x + \Delta x) - g(x)}{\Delta x}\right)\right] \]
\[ = \left[\lim_{\Delta x \to 0} g(x)\right]\left[\lim_{\Delta x \to 0} \frac{h(x + \Delta x) - h(x)}{\Delta x}\right] + \left[\lim_{\Delta x \to 0} h(x + \Delta x)\right]\left[\lim_{\Delta x \to 0}\frac{g(x + \Delta x) - g(x)}{\Delta x}\right] \]
\[ = g(x)h'(x) + h(x)g'(x) \]