In the Problem section, we found out that the derivative of \((ax+b)(cx+d)\) is \(a(cx+d)+c(ax+b)\). It turns out that we can use this result to differentiate the product of *any* two functions, and in this section, we explore how to do so.

Our key question is:

What is the derivative of the product \(h(x)=f(x)g(x)\) for any functions \(f\) and \(g\)?

We will find the derivative of \(h(x)\) at the point \(x=x_0\).

In A tangent is…, we saw that we can locally approximate any function by a straight line (so long as we avoid any points where the gradient is not well defined). So near \(x_0\), \(f(x)\approx ax+b\) for some \(a\) and \(b\).

We can find \(a\) and \(b\) by matching the gradients and \(y\)-values: \[f'(x_0)=a \quad\text{and}\quad f(x_0)=ax_0+b\] Likewise, near \(x_0\), \(g(x)\approx cx+d\) where \(c\) and \(d\) are such that \(g'(x_0)=c\) and \(g(x_0)=cx_0+d\).

So near \(x_0\), we have \(h(x)=f(x)g(x)\approx (ax+b)(cx+d)\), so using the rule we found out, we see that the derivative of \(h(x)\) at \(x_0\) is \(a(cx_0+d)+c(ax_0+b)\). Substituting in our values for \(a\), \(b\), \(c\) and \(d\), we find \[h'(x_0)=f'(x_0)g(x_0)+g'(x_0)f(x_0).\] So in general, \[h'(x)=f'(x)g(x)+g'(x)f(x).\]

An alternative notation which is commonly used is to write \(u=f(x)\) and \(v=g(x)\). Then we have \[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}.\]