7.3.4 Composition of Functions

Find the derivative of the function $f:ℝ\to ℝ$ 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 $v\text{'}(u)=5u^4$. Inserting these terms into the chain rule results in the derivative $f\text{'}$ of the function $f$ with

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

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

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