Chapter 3.4, Problem 1E

Write the composite function in the form f(g(x)). [Identify the inner function u = g(x) and the outer function y = f(u).] Then find the derivative dy/dx.

$y=\sqrt[3]{1+4x}$

To determine

To find: The composite function in the form $f\left(g\left(x\right)\right)$ and obtain the derivative of y.

Answer to Problem 1E

The inner function is $u=1+4x$ and the outer function is $f\left(u\right)=\sqrt[3]{u}$.

The derivative of y is $\frac{dy}{dx}=\frac{4}{3\sqrt[3]{{\left(1+4x\right)}^2}}$.

Explanation of Solution

Given:

The function is $y=\sqrt[3]{1+4x}$.

Formula used:

The Chain Rule:

If h is differentiable at x and g is differentiable at $h\left(x\right)$, then the composite function $F=g\circ h$ defined by $F\left(x\right)=g\left(h\left(x\right)\right)$ is differentiable at x and $F^{\prime }$ is given by the product,

$F^{\prime }\left(x\right)=g^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (1)

Derivative Rule:

(1) Power Rule: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

(2) Sum Rule: $\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]=\frac{d}{dx}\left[f\left(x\right)\right]+\frac{d}{dx}\left[g\left(x\right)\right]$

Calculation:

Let the inner function be $u=g\left(x\right)$ and the outer function be $y=f\left(u\right)$.

Then, $g\left(x\right)=1+4x$ and $f\left(u\right)=\sqrt[3]{u}$. That is,

$\begin{array}{c}y=\sqrt[3]{1+4x}\\ =f\left(1+4x\right)\\ =f\left(g\left(x\right)\right)\end{array}$

Therefore, $y=f\left(g\left(x\right)\right)$.

Hence, the inner function is $u=1+4x$ and the outer function is $f\left(u\right)=\sqrt[3]{u}$.

Thus, the required form of composite function is $f\left(g\left(x\right)\right)=\sqrt[3]{u}$.

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left(y\right)\\ =\frac{d}{dx}\left(\sqrt[3]{1+4x}\right)\end{array}$

Let $h\left(x\right)=1+4x$ and $g\left(u\right)={\left(u\right)}^{\frac{1}{3}}\text{ where }u=h\left(x\right)$.

Apply the chain rule as shown in equation (1),

$F^{\prime }\left(x\right)=g^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (2)

The derivative $g^{\prime }\left(h\left(x\right)\right)$ is computed as follows,

$\begin{array}{c}{g}^{\prime }\left(h\left(x\right)\right)=g^{\prime }\left(u\right)\\ =\frac{d}{du}\left(g\left(u\right)\right)\\ =\frac{d}{du}{\left(u\right)}^{\frac{1}{3}}\end{array}$

Apply the power rule (1),

$\begin{array}{c}{g}^{\prime }\left(h\left(x\right)\right)=\left(\frac{1}{3}{u}^{\frac{1}{3}-1}\right)\\ =\frac{1}{3}{\left(u\right)}^{\frac{1-3}{3}}\\ =\frac{1}{3}{\left(u\right)}^{\frac{-2}{3}}\\ =\frac{1}{3{\left(u\right)}^{\frac{2}{3}}}\end{array}$

Substitute $u=1+4x$ in above equation.

$g^{\prime }\left(h\left(x\right)\right)=\frac{1}{3{\left(1+4x\right)}^{\frac{2}{3}}}$

The derivative $g^{\prime }\left(h\left(x\right)\right)$ is $g^{\prime }\left(h\left(x\right)\right)=\frac{1}{3{\left(1+4x\right)}^{\frac{2}{3}}}$.

The derivative of $h\left(x\right)$ is computed as follows,

$h^{\prime }\left(x\right)=\frac{d}{dx}\left(1+4x\right)$

Apply sum rule (2) and the power rule (1),

$\begin{array}{c}{h}^{\prime }\left(x\right)=\frac{d}{dx}\left[1+4x\right]\\ =\frac{d}{dx}\left[1\right]+\frac{d}{dx}\left[4x\right]\\ =0+4\frac{d}{dx}\left[x\right]\\ =4\left[1x^{1-1}\right]\\ =4\end{array}$

Thus, the derivative $h\left(x\right)$ is $h^{\prime }\left(x\right)=4$.

Substitute $\frac{1}{3{\left(1+4x\right)}^{\frac{2}{3}}}$ for $g^{\prime }\left(h\left(x\right)\right)$ and $4$ for $h^{\prime }\left(x\right)$ in equation (2).

$\begin{array}{c}{g}^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)=\frac{1}{3{\left(1+4x\right)}^{\frac{2}{3}}}\left(4\right)\\ =\frac{4}{3\cdot \sqrt[3]{{\left(1+4x\right)}^2}}\end{array}$

Therefore, the derivative of $y=\sqrt[3]{1+4x}$ is $y^{\prime }=\frac{4}{3\cdot \sqrt[3]{{\left(1+4x\right)}^2}}$.