The function $F\left(x\right)\={\left(x^2\+1\right)}^3$ is the composition of the functions $f\left(x\right)\=x^3$ and $g\left(x\right)\=x^2+1\,$ so that $f\left(x\right)\=f\left(g\left(x\right)\right)$. Thus, $F\left(x\right)$ is called a composite function.

The number $F\left(2\right)$ is computed by substituting $x\=2$ into $x^2+1$ to get 5, then cubing 5 to get 125. The first function that must be evaluated will be called the "inner" function, while the second function that gets applied afterwards will be called the "outer" function. Although $g\left(x\right)\=x^2+1$ can itself be thought of as a composite function, in the context of differentiation that will usually not be necessary. What is crucial for differentiation is the correct recognition of the outer and inner functions.

Using purely functional notation, the composition of the functions $f$ and $g$ would be written $F\=f\circ g$. Using this notation, the value of the composite function $F$ at $x$ is expressed by the notation $F\(x\)\=\(f\circ g\)\(x\)$. The notation $F\(x\)\=f\(g\(x\)\)$ generally proves to be simpler to use. (The functional notation $f\circ g$ looks much like the word "fog" and often causes it.)

Additional examples of composite functions appear in Table 2.4.2.

(The word composition has but one pronunciation; the word composite, two. It is pronounced com-POS'-it in the USA, and COMP'-o-zit in other parts of the English-speaking world.)

Differentiation of a composite function generally requires the chain rule$\.$  Of all the rules of differentiation, the chain rule is the most important, more for theoretical reasons than for the immediate need to evaluate derivatives of specific functions.

The phrase "the derivative of the stuff inside" is probably the most important mnemonic that can be attached to the study of the chain rule.

Theorem 2.4.1 is a formal statement of the chain rule.

One way to think of the Chain rule is to think of composition as a gear. For example, if a chain is being pulled through a 53-tooth gear and if the gear is being turned at a rate of 100 revolutions per minute then the chain passes through the gear at a rate of (53 links/revolution)(100 revolutions/min) = 5300 links/min. Note how the composite rate is the product of the individual rates.

The function $g$ is the (functional) inverse of the function $f$ if the compositions $f\left(g\left(x\right)\right)$ and $g\left(f\left(x\right)\right)$ both simplify to $x$. The typical notation for the function inverse to $f$ is $f^{-1}$, so the compositions that must hold are $f\left(f^{-1}\left(x\right)\right)\=x\=f^{-1}\left(f\left(x\right)\right)$.

Let $g$ be the functional inverse of $f$. The derivative of $g$ is given by the Inverse-Function rule.

Thus, the derivatives of inverse functions are related as reciprocals, but the derivative of the reciprocated function must be evaluated at the inverse function. This rule follows easily from the Chain rule and the identity $f\left(g\left(x\right)\right)\=x$ that holds when $f$ and $g$ are inverse functions. The Chain rule gives $f\prime \left(g\left(x\right)\right)\cdot g\prime \left(x\right)\=1$, from which the reciprocal rule follows by division.$$

If a curve $C$ is defined parametrically by $x\=x\left(t\right)\,y\=y\left(t\right)$, the slope along the curve is given by

$\frac{d}{\mathrm{dx}}\stackrel{\^}{y}\left(x\right)\=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$ 

where $\stackrel{\^}{y}\left(x\right)\=y\left(t\left(x\right)\right)$. Unfortunately, nearly all texts fail to distinguish $\stackrel{\^}{y}$ from $y$, even though these are completely different functions. When this notational omission is made, the formula for the derivative of $\stackrel{\^}{y}$ assumes the perplexing form

$\frac{\mathrm{dy}}{\mathrm{dx}}\=\frac{y\prime \left(t\right)}{x\prime \left(t\right)}$ 

The most appropriate form for this formula would be the very explicit

$\frac{d}{\mathrm{dx}}\stackrel{\^}{y}\left(x\right)\=\genfrac{}{}{0ex}{}{\frac{y\prime \left(t\right)}{x\prime \left(t\right)}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{t\=t\left(x\right)}$ 

which follows from an application of the Chain and Inverse-Function rules for derivatives.

$\frac{d}{\mathrm{dx}}\stackrel{\^}{y}\left(x\right)\=\frac{d}{\mathrm{dx}}y\left(t\left(x\right)\right)\=\frac{\mathrm{dy}}{\mathrm{dt}}\frac{\mathrm{dt}}{\mathrm{dx}}\=\frac{\mathrm{dy}}{\mathrm{dt}}\left(\frac{1}{\frac{\mathrm{dx}}{\mathrm{dt}}}\right)\=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}\=\frac{y\prime \left(t\right)}{x\prime \left(t\right)}$ 

The prime symbol used for differentiation represents an operator. By default, Maple interprets the prime to be the operator $\frac{d}{\mathrm{dx}}$. If the differentiation variable is other than $x$, then one way to have Maple re-interpret the prime is to include the independent variable explicitly. So, if using the prime to obtain the ratio of derivatives, include the appropriate independent variable as an explicit argument.

As a final caution, note that shortening the differentiation rule to something like $y\prime \=y\prime \/x\prime$ would befuddle even the most astute reader.

Table 2.4.3 summarizes the key ideas in Section 2.4.