Chapter 3.4, Problem 62E

To determine

To find: The derivative $f^{″}\left(x\right)$ in terms of $g,g^{\prime }\text{ and }{g}^{″}$.

Answer to Problem 62E

The derivative $f^{″}\left(x\right)$ in terms of $g,g^{\prime }\text{ and }{g}^{″}$ is $f^{″}\left(x\right)=4x^3{g}^{″}\left(x^2\right)+6x{g}^{\prime }\left(x^2\right)$.

Explanation of Solution

Given:

The function is $f\left(x\right)=xg\left(x^2\right)$.

Result used: 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 Rules:

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

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:

Obtain the first derivative of $f\left(x\right)=xg\left(x^2\right)$.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(f\left(x\right)\right)\\ =\frac{d}{dx}\left(xg\left(x^2\right)\right)\end{array}$

Apply the product rule (1),

$\begin{array}{c}{f}^{\prime }\left(x\right)=x\frac{d}{dx}\left(g\left(x^2\right)\right)+g\left(x^2\right)\frac{d}{dx}\left(x\right)\\ =x\frac{d}{dx}\left(g\left(x^2\right)\right)+g\left(x^2\right)\left(1x^{1-1}\right)\end{array}$

$=x\frac{d}{dx}\left(g\left(x^2\right)\right)+g\left(x^2\right)$ (2)

Obtain the derivative $\frac{d}{dx}\left(g\left(x^2\right)\right)$ by using the chain rule as shown in equation (1).

Let $h\left(x\right)=x^2$ and $f\left(u\right)=g\left(u\right)\text{ where }u=h\left(x\right)$.

$\frac{d}{dx}\left(g\left(x^2\right)\right)=g^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (3)

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

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

Substitute $u=x^2$ in the above equation,

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

Thus, the derivative is $f^{\prime }\left(h\left(x\right)\right)=g^{\prime }\left(x^2\right)$.

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

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

Thus, the derivative is $h^{\prime }\left(x\right)=2x$.

Substitute $g^{\prime }\left(x^2\right)$ for $f^{\prime }\left(h\left(x\right)\right)$ and $2x$ for $h^{\prime }\left(x\right)$ in equation (3),

$\begin{array}{c}\frac{d}{dx}\left(g\left(x^2\right)\right)=g^{\prime }\left(x^2\right)\cdot 2x\\ =2x\cdot g^{\prime }\left(x^2\right)\end{array}$

Substitute $2x\cdot g^{\prime }\left(x^2\right)$ for $\frac{d}{dx}\left(g\left(x^2\right)\right)$ in equation (2),

$\begin{array}{c}{f}^{\prime }\left(x\right)=x\left(2x\cdot g^{\prime }\left(x^2\right)\right)+g\left(x^2\right)\\ =2x^2{g}^{\prime }\left(x^2\right)+g\left(x^2\right)\end{array}$

Therefore, the derivative of $f\left(x\right)=xg\left(x^2\right)$ is $f^{\prime }\left(x\right)=2x^2{g}^{\prime }\left(x^2\right)+g\left(x^2\right)$.

Obtain the second derivative of $f\left(x\right)$.

$\begin{array}{c}{f}^{″}\left(x\right)=\frac{d}{dx}\left(f^{\prime }\left(x\right)\right)\\ =\frac{d}{dx}\left(2x^2{g}^{\prime }\left(x^2\right)+g\left(x^2\right)\right)\end{array}$

Apply the sum rule (2),

$f^{″}\left(x\right)=\frac{d}{dx}\left(2x^2{g}^{\prime }\left(x^2\right)+g\left(x^2\right)\right)$

$=\frac{d}{dx}\left(2x^2{g}^{\prime }\left(x^2\right)\right)+\frac{d}{dx}\left(g\left(x^2\right)\right)$ (4)

Obtain the derivative $\frac{d}{dx}\left(2x^2{g}^{\prime }\left(x^2\right)\right)$ by using the product rule (1),

$\begin{array}{c}\frac{d}{dx}\left(2x^2{g}^{\prime }\left(x^2\right)\right)=2x^2\frac{d}{dx}\left[g^{\prime }\left(x^2\right)\right]+g^{\prime }\left(x^2\right)\frac{d}{dx}\left[2x^2\right]\\ =2x^2\frac{d}{dx}\left[g^{\prime }\left(x^2\right)\right]+g^{\prime }\left(x^2\right)2\left[2x^{2-1}\right]\end{array}$

$=2x^2\frac{d}{dx}\left[g^{\prime }\left(x^2\right)\right]+g^{\prime }\left(x^2\right)\left[4x\right]$ (5)

Obtain the derivative $\frac{d}{dx}\left(g^{\prime }\left(x^2\right)\right)$ by using the chain rule as shown in equation (1).

Let $h\left(x\right)=x^2$ and $k\left(u\right)=g^{\prime }\left(u\right)\text{ where }u=h\left(x\right)$.

$\frac{d}{dx}\left(g^{\prime }\left(x^2\right)\right)=k^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (6)

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

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

Substitute $u=x^2$ in the above equation,

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

Thus the derivative is $g^{\prime }\left(h\left(x\right)\right)=g^{″}\left(x^2\right)$.

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

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

Thus the derivative is $h^{\prime }\left(x\right)=2x$.

Substitute $g^{″}\left(x^2\right)$ for $k^{\prime }\left(h\left(x\right)\right)$ and $2x$ for $h^{\prime }\left(x\right)$ in equation (6),

$\begin{array}{c}\frac{d}{dx}\left(g^{\prime }\left(x^2\right)\right)=g^{″}\left(x^2\right)\cdot 2x\\ =2x\cdot g^{″}\left(x^2\right)\end{array}$

Substitute $2x\cdot g^{″}\left(x^2\right)$ for $\frac{d}{dx}\left(g^{\prime }\left(x^2\right)\right)$ in equation (5),

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

Substitute $4x^3{g}^{″}\left(x^2\right)+4x{g}^{\prime }\left(x^2\right)$ for $\frac{d}{dx}\left(2x^2{g}^{\prime }\left(x^2\right)\right)$ and $2x\cdot g^{\prime }\left(x^2\right)$ for $\frac{d}{dx}\left(g\left(x^2\right)\right)$ in equation (4),

$\begin{array}{c}{f}^{″}\left(x\right)=4x^3{g}^{″}\left(x^2\right)+4x{g}^{\prime }\left(x^2\right)+2x\cdot g^{\prime }\left(x^2\right)\\ =4x^3{g}^{″}\left(x^2\right)+6x{g}^{\prime }\left(x^2\right)\end{array}$

Therefore, the derivative $f^{″}\left(x\right)$ in terms of $g,g^{\prime }\text{ and }{g}^{″}$ is

$f^{″}\left(x\right)=4x^3{g}^{″}\left(x^2\right)+6x{g}^{\prime }\left(x^2\right)$.