Chapter 3.4, Problem 56E

(a)

To determine

To find: The value $h^{\prime }\left(2\right)$.

Answer to Problem 56E

The value $h^{\prime }\left(2\right)$ is approximately 1.

Explanation of Solution

Given:

The function is $h\left(x\right)=f\left(f\left(x\right)\right)$.

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)

The slope of the line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$.

Calculation:

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

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

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

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

Substitute $x=2$ in the above equation,

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

From the given graph it is observed that, $f\left(2\right)=1$.

$h^{\prime }\left(2\right)=f^{\prime }\left(1\right)f^{\prime }\left(2\right)$ (2)

Obtain the values $f^{\prime }\left(1\right)$ and $f^{\prime }\left(2\right)$.

From the given graph, draw the tangent line at $x=1$ is approximately joining the points $\left(1,\left(f\left(1\right)\right)\right)$ and $\left(0.9,\left(f\left(0.9\right)\right)\right)$.

Use the slope formula stated above and compute the slope of the line passing through the points $\left(1,f\left(1\right)\right)$ and $\left(0.9,\left(f\left(0.9\right)\right)\right)$.

$f^{\prime }\left(1\right)=\frac{f\left(0.9\right)-f\left(1\right)}{0.9-1}$

From the given graph it is observed that,$f\left(0.9\right)=2.3$ and $f\left(1\right)=2.2$.

$\begin{array}{c}{f}^{\prime }\left(1\right)=\frac{2.3-2.2}{0.9-1}\\ =\frac{0.1}{-0.1}\\ =-1\end{array}$

Thus, the value is $f^{\prime }\left(1\right)=-1$.

Similarly, the value $f^{\prime }\left(2\right)$ is $f^{\prime }\left(2\right)=-1$.

Substitute $-1$ for $f^{\prime }\left(1\right)$ and $-1$ for $f^{\prime }\left(2\right)$ in equation (2),

$\begin{array}{c}{h}^{\prime }\left(1\right)\approx \left(-1\right)\left(-1\right)\\ =1\end{array}$

Therefore, the value $h^{\prime }\left(2\right)$ is approximately 1.

(b)

To determine

To find: The value $g^{\prime }\left(2\right)$.

Answer to Problem 56E

The value $g^{\prime }\left(2\right)$ is approximately 8.

Explanation of Solution

Given:

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

Calculation:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Substitute $x=2$ in the above equation,

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

$=4\cdot f^{\prime }\left(4\right)$ (4)

Obtain the value of $f^{\prime }\left(4\right)$.

From the given graph, draw the tangent line at $x=4$ is approximately joining the points $\left(4,\left(f\left(4\right)\right)\right)$ and $\left(4.1,\left(f\left(4.1\right)\right)\right)$.

Use the slope formula stated above and compute the slope of the line passing through the points $\left(4,\left(f\left(4\right)\right)\right)$ and $\left(4.1,\left(f\left(4.1\right)\right)\right)$.

$f^{\prime }\left(4\right)=\frac{f\left(4.1\right)-f\left(4\right)}{4.1-4}$

From the given graph observe that,$f\left(4\right)=2$ and $f\left(4.1\right)=2.2$.

$\begin{array}{c}{f}^{\prime }\left(4\right)=\frac{2.2-2}{4.1-4}\\ =\frac{0.2}{0.1}\\ =2\end{array}$

Thus, the value $f^{\prime }\left(4\right)$ is $f^{\prime }\left(4\right)=2$.

Substitute 2 for $f^{\prime }\left(4\right)$ in equation (4),

$\begin{array}{c}{g}^{\prime }\left(2\right)\approx 4\cdot 2\\ =8\end{array}$

Therefore, the value $g^{\prime }\left(2\right)$ is approximately 8.