#### To determine

**To find:** The value F′(2).

#### Answer

The value F′(2) is F′(2)=20_.

#### Explanation

**Given:**

The function is F(x)=f(f(x)).

**Result used: Chain Rule:**

If *h* is differentiable at *x* and *g* is differentiable at h(x), then the composite function F=g∘h defined by F(x)=g(h(x)) is differentiable at *x* and F′ is given by the product,

F′(x)=g′(h(x))⋅h′(x) (1)

**Calculation:**

Obtain the derivative of F(x)=f(f(x)).

F′(x)=ddx(F(x))=ddx(f(f(x)))

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

F′(x)=f′(f(x))⋅f′(x) (2)

Substitute x=2 in equation (2),

F′(2) =f′(f(2))⋅f′(2)

Consider the values from given table in exercise (63), f(2)=1 and f′(2)=5.

F′(2) =f′(1)⋅5

Consider the values from given table in exercise (63), f′(1)=4.

F′(2) =4⋅5=20

Therefore, the value F′(2) is F′(2)=20_.

#### To determine

**To find:** The value G′(3).

#### Answer

The value G′(3) is G′(3)=63_.

#### Explanation

**Given:**

The function is G(x)=g(g(x)).

**Calculation:**

Obtain the derivative of G(x)=g(g(x)).

G′(x)=ddx(G(x))=ddx(g(g(x)))

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

G′(x)=g′(g(x))⋅g′(x) (3)

Substitute x=3 in equation (2),

G′(3) =g′(g(3))⋅g′(3)

Consider the values from given table in exercise (63), g(3)=2 and g′(3)=9 .

G′(3) =g′(2)⋅9

Consider the values from given table in exercise (63), g′(2)=7.

G′(3) =7⋅9=63

Therefore, the value G′(3) is G′(3)=63_.