Differentiating products:

Suppose that $h:(-\infty,\infty)\rightarrow(-\infty,\infty)$ is given by $h(x) = x^2\sin(x)$. None of our formulae to date would let us differentiate this function, yet we recognize that it is the product of two functions which we do know how to differentiate.

Consider the general situation. We are given $f:D\rightarrow(-\infty,\infty)$and $g:D\rightarrow(-\infty,\infty)$, and that both f'(a) and g'(a) exist. (Remember that f'(a) and g'(a) are limits!) If we define $f\cdot g:D\rightarrow(-\infty,\infty)$ by $(f\cdot g)(x) = f(x)g(x)$, how do we compute $(f\cdot g)'(a)$? Observe that

$$\begin{array} {rcl} (f\cdot g)(a+h) - (f\cdot g)(a) & = & f(... ...a)) + (f(a+h)-f(a))(g(a+h)-g(a)) + g(a)(f(a+h)-f(a))\end{array}$$

so

$$\begin{array} {rcl} \frac{(f\cdot g)(a+h) - (f\cdot g)(a)}{a... ...h)-g(a)}{h}\right)\\ &&+\;g(a)\frac{f(a+h)-f(a)}{h}.\end{array}$$

Notice that each factor of each term on the right side of the preceding equation has a limit as h approaches . For example, the first quotient converges to g'(a). Therefore

$$\begin{array} {rcl} (f\cdot g)'(a) & = & \lim_{h\rightarrow ... ...f'(a)g'(a) + g(a)f'(a)\\ & = & f(a)g'(a) + f'(a)g(a)\end{array}$$

So, for example, we see in our example where $h(x) = x^2\sin(x)$ that $h'(x) = x^2\cos(x) + 2x\sin(x)$.

Here is an important example: Suppose that $h:(0,\infty)\rightarrow(-\infty,\infty)$ with $h(x) = x(-1 + \ln(x))$. Then

$$\begin{array} {rcl} h'(x) & = & x\left(0 + \frac{1}{x}\right) + 1(-1 + \ln(x))\\ & = & \ln(x)\end{array}$$

Notice that we have produced a function whose derivative is $\ln$, and this function is not $\ln$ itself!! The only functions that are there own derivatives are functions whose rule is of the form f(x) = Ae^x, where A is a constant!!!!