#### Chapter 7 Differential Calculus

Section 7.3 Calculation Rules

# 7.3.4 Composition of Functions

Finally, we investigate composition of functions (see Module 6, Section 6.6.3): what happens if a function $u$ (the inner function) is substituted into another function $v$ (the outer function)? In mathematics, such a composition is denoted by $f:=v\circ u$ with $f\left(x\right)=\left(v\circ u\right)\left(x\right):=v\left(u\left(x\right)\right)$. That is, first the value of a function $u$ is determined depending on the variable $x$. The value $u\left(x\right)$ calculated this way is then used as an argument of the function $v$. This results in the final function value $v\left(u\left(x\right)\right)$.
##### Chain Rule 7.3.7
The derivative of the function $f:=v\circ u$ with $f\left(x\right)=\left(v\circ u\right)\left(x\right):=v\left(u\left(x\right)\right)$ can be calculated applying the chain rule:

$f\text{'}\left(x\right)=v\text{'}\left(u\left(x\right)\right)·u\text{'}\left(x\right) .$

Here, the expression $v\text{'}\left(u\left(x\right)\right)$ is considered in such a way that $v$ is a function of $u$ and thus, the derivative is taken with respect to $u$; then $v\text{'}\left(u\right)$ is evaluated for $u=u\left(x\right)$.
The following phrase is a useful summary: the derivative of a composite function is the product of the outer derivative and the inner derivative.

This differentiation rule shall be illustrated by a few examples.
##### Example 7.3.8
Find the derivative of the function $f:ℝ\to ℝ$ with $f\left(x\right)=\left(3-2x{\right)}^{5}$. To apply the chain rule, inner and outer functions must be identified. If we take the function $u\left(x\right)=3-2x$ as the inner function $u$, then the outer function $v$ is given by $v\left(u\right)={u}^{5}$. With this, we have the required form $v\left(u\left(x\right)\right)=f\left(x\right)$.
Taking the derivative of the inner function $u$ with respect to $x$ results in $u\text{'}\left(x\right)=-2$. For the outer derivative, the function $v$ is differentiated with respect to $u$, which results in $v\text{'}\left(u\right)=5{u}^{4}$. Inserting these terms into the chain rule results in the derivative $f\text{'}$ of the function $f$ with

$f\text{'}\left(x\right)=5\left(u\left(x{\right)\right)}^{4}·\left(-2\right)=5\left(3-2x{\right)}^{4}·\left(-2\right)=-10\left(3-2x{\right)}^{4} .$

As a second example, let's calculate the derivative of $g:ℝ\to ℝ$ with $g\left(x\right)=e{}^{{x}^{3}}$. For the inner function $u$ the assignment $x\to u\left(x\right)={x}^{3}$ and for the outer function $v$ the assignment $u\to v\left(u\right)=e{}^{u}$ is appropriate. Taking the inner and the outer derivative results in $u\text{'}\left(x\right)=3{x}^{2}$ and $v\text{'}\left(u\right)=e{}^{u}$. Inserting these terms into the chain rule results in the derivative of the function $g$:

$g\text{'}:ℝ\to ℝ , x\to g\text{'}\left(x\right)=e{}^{u\left(x\right)}·3{x}^{2}=e{}^{{x}^{3}}·3{x}^{2}=3{x}^{2}e{}^{{x}^{3}} .$