Composite functions:

Suppose that $f:D\rightarrow(-\infty,\infty)$,$g:E\rightarrow(-\infty,\infty)$ and the range of f is contained in E, the domain of g. We can then define a new function $g\circ f:D\rightarrow(-\infty,\infty)$ by the rule $(g\circ f)(x) = g(f(x))$. $g\circ f$ is called the composition of g with f. We wish to find the derivative of $g\circ f$ if we have information about the derivatives of f and g separately. To see what we need to know, we will reason somewhat informally, and not worry to much about if we might divide by , or commit other sins.

$$\begin{array} {rcl} \frac{(g\circ f)(a+h) - (g\circ f)(a)}{a... ...g(f(a+h))-g(f(a))}{f(a+h)-f(a)}\frac{f(a+h)-f(a)}{h}\end{array}$$

This is progress since we recognize that

$$\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h} = f'(a).$$

The problem is what is

$$\lim_{h\rightarrow 0}\frac{g(f(a+h)) - g(f(a))}{f(a+h) - f(a)}?$$

We do recognize that it is the slope of the chord on the graph of y = g(x) drawn through the points (f(a), g(f(a))) and (f(a+h), g(f(a+h))), presuming these points are different! Ignoring this pitfall for the moment, since we know that $\lim_{h\rightarrow 0}f(a+h) = f(a)$, we would believe that the

$$\lim_{h\rightarrow 0}\frac{g(f(a+h)) - g(f(a))}{f(a+h) - f(a... ...im_{x\rightarrow f(a)}\frac{g(x) - g(f(a))}{x-f(a)} = g'(f(a)),$$

so that

$$\begin{array} {rcl} \lim_{h\rightarrow 0}\frac{(g\circ f)(a+... ...rac{f(a+h)-f(a)}{h}\right)\\ & ``='' & g'(f(a))f'(a)\end{array}$$

Well, it turns out that this heuristic calculation leads to the right answer, but it is a little tricky to see why. We will postpone the justification to the next lecture. Here are some examples of the formula in use.

$H:(-\infty,\infty)\rightarrow(-\infty,\infty)$, H(x) = e^-x. Then H'(x) = e^-x(-1).

$H:(-\infty,\infty)\rightarrow(-\infty,\infty)$, $H(x) = \sin(x^2)$. Then $H'(x) = \cos(x^2)\cdot(2x)$.

Here is an example in combination with the product rule: $H:(-\infty,\infty)\rightarrow(-\infty,\infty)$, $H(x) = x\arctan(x) - (1/2)\ln(x^2+1)$.

$$\begin{array} {rcl} H'(x) & = & x\frac{1}{1+x^2} + (1)\arctan(x) - (1/2)\frac{1}{1+x^2}(0+2x)\\ & = & \arctan(x)\end{array}$$

Note that H is a rather complicated function with a rather simple derivative.

Here is a similar example: $H:[-1,1]\rightarrow(-\infty,\infty)$ with $H(x) = x\arcsin(x) + \sqrt{1-x^2}$.Note that we do not expect H to have a derivative at $\pm 1$ since neither $\arcsin$ does not have a derivative at $\pm 1$ and $\sqrt{}$ does not have a derivative at $1 - (\pm 1)^2 = 0$. Remember that $\sqrt{1-x^2} = (1-x^2)^{1/2}$. So for $x\in (-1,1)$ we have

$$\begin{array} {rcl} H'(x) & = & {\displaystyle x\frac{1}{\s... ...csin(x) - \frac{x}{\sqrt{1-x^2}}}\\ & = & \arcsin(x)\end{array}$$

Since $\arcsin(1) = \pi/2$, we might wonder if H is differentiable at 1. The chain rule will not help us! The only way to answer this question is to try to see if (H(1 + h) - H(1))/h has a limit as h approaches 1 from below.