Chapter 3.4, Problem 54E

Let f and g be the functions in Exercise 63.

(a) If F(x) = f(f(x)), find F′(2).

(b) If G(x) = g(g(x)), find G′(3).

To determine

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

Answer to Problem 54E

The value $F^{\prime }\left(2\right)$ is $\underset{\_}{F^{\prime }\left(2\right)=20}$.

Explanation of Solution

Given:

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

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)

Calculation:

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

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

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

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

Substitute $x=2$ in equation (2),

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

Consider the values from given table in exercise (63), $f\left(2\right)=1$ and $f^{\prime }\left(2\right)=5$.

$F^{\prime }\left(2\right)=f^{\prime }\left(1\right)\cdot 5$

Consider the values from given table in exercise (63), $f^{\prime }\left(1\right)=4$.

$\begin{array}{c}{F}^{\prime }\left(2\right)=4\cdot 5\\ =20\end{array}$

Therefore, the value $F^{\prime }\left(2\right)$ is $\underset{\_}{F^{\prime }\left(2\right)=20}$.

To determine

To find: The value $G^{\prime }\left(3\right)$.

Answer to Problem 54E

The value $G^{\prime }\left(3\right)$ is $\underset{\_}{G^{\prime }\left(3\right)=63}$.

Explanation of Solution

Given:

The function is $G\left(x\right)=g\left(g\left(x\right)\right)$.

Calculation:

Obtain the derivative of $G\left(x\right)=g\left(g\left(x\right)\right)$.

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

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

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

Substitute $x=3$ in equation (2),

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

Consider the values from given table in exercise (63), $g\left(3\right)=2$ and $g^{\prime }\left(3\right)=9$ .

$G^{\prime }\left(3\right)=g^{\prime }\left(2\right)\cdot 9$

Consider the values from given table in exercise (63), $g^{\prime }\left(2\right)=7$.

$\begin{array}{c}{G}^{\prime }\left(3\right)=7\cdot 9\\ =63\end{array}$

Therefore, the value $G^{\prime }\left(3\right)$ is $\underset{\_}{G^{\prime }\left(3\right)=63}$.