#### 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 $f(x)=(v\circ u)(x):=v(u(x))$. That is, first the value of a function $u$ is determined depending on the variable $x$. The value $u(x)$ calculated this way is then used as an argument of the function $v$. This results in the final function value $v(u(x))$.

##### **Chain Rule 7.3.7 **

The derivative of the function $f:=v\circ u$ with $f(x)=(v\circ u)(x):=v(u(x))$ can be calculated applying the

$f\text{'}(x)=v\text{'}(u(x))\xb7u\text{'}(x)\hspace{0.5em}.$

Here, the expression $v\text{'}(u(x))$ is considered in such a way that $v$ is a function of $u$ and thus, the derivative is taken with respect to $u$; then $v\text{'}(u)$ is evaluated for $u=u(x)$.

The following phrase is a useful summary: the derivative of a composite function is the product of the outer derivative and the inner derivative.

**chain rule**:Here, the expression $v\text{'}(u(x))$ is considered in such a way that $v$ is a function of $u$ and thus, the derivative is taken with respect to $u$; then $v\text{'}(u)$ is evaluated for $u=u(x)$.

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 $f:\mathbb{R}\to \mathbb{R}$ with $f(x)=(3-2x{)}^{5}$. To apply the chain rule, inner and outer functions must be identified. If we take the function $u(x)=3-2x$ as the inner function $u$, then the outer function $v$ is given by $v(u)={u}^{5}$. With this, we have the required form $v(u(x))=f(x)$.

Taking the derivative of the inner function $u$ with respect to $x$ results in $u\text{'}(x)=-2$. For the outer derivative, the function $v$ is differentiated with respect to $u$, which results in $v\text{'}(u)=5{u}^{4}$. Inserting these terms into the chain rule results in the derivative $f\text{'}$ of the function $f$ with

$f\text{'}(x)=5(u(x{))}^{4}\xb7(-2)=5(3-2x{)}^{4}\xb7(-2)=-10(3-2x{)}^{4}\hspace{0.5em}.$

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

$g\text{'}:\mathbb{R}\to \mathbb{R}\hspace{0.5em},\hspace{0.5em}\hspace{0.5em}x\to g\text{'}(x)=e{}^{u(x)}\xb73{x}^{2}=e{}^{{x}^{3}}\xb73{x}^{2}=3{x}^{2}e{}^{{x}^{3}}\hspace{0.5em}.$

Taking the derivative of the inner function $u$ with respect to $x$ results in $u\text{'}(x)=-2$. For the outer derivative, the function $v$ is differentiated with respect to $u$, which results in $v\text{'}(u)=5{u}^{4}$. Inserting these terms into the chain rule results in the derivative $f\text{'}$ of the function $f$ with

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