#### To determine

**To find:** The value h′(2).

#### Answer

The value h′(2) is approximately 1.

#### Explanation

**Given:**

The function is h(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)

The slope of the line passing through the points (x1,y1) and (x2,y2) is y2−y1x2−x1.

**Calculation:**

Obtain the derivative of h(x)=f(f(x))*.*

h′(x)=ddx(h(x))=ddx(f(f(x)))

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

h′(x)=f′(f(x))f′(x)

Substitute x=2 in the above equation,

h′(2)=f′(f(2))f′(2)

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

h′(2)=f′(1)f′(2) (2)

Obtain the values f′(1) and f′(2).

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

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

f′(1)=f(0.9)−f(1)0.9−1

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

f′(1)=2.3−2.20.9−1=0.1−0.1=−1

Thus, the value is f′(1)=−1.

Similarly, the value f′(2) is f′(2)=−1.

Substitute −1 for f′(1) and −1 for f′(2) in equation (2),

h′(1)≈(−1)(−1)=1

Therefore, the value h′(2) is approximately 1.

#### To determine

**To find:** The value g′(2).

#### Answer

The value g′(2) is approximately 8.

#### Explanation

**Given:**

The function is g(x)=f(x2).

**Calculation:**

Obtain the derivative of g(x)=f(x2)*.*

g′(x)=ddx(g(x))=ddx(f(x2))

Let h(x)=x2 and g(u)=f(u) where u=h(x).

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

g′(x)=g′(h(x))⋅h′(x) (3)

The derivative of g′(h(x)) is computed as follows,

g′(h(x))=g′(u)=ddu(g(u))=ddu(f(u))=f′(u)

Substitute u=x2 in the above equation,

g′(h(x))=f′(x2)

Thus, the derivative is g′(h(x))=f′(x2).

The derivative of h(x) is computed as follows,

h′(x)=ddx(x2)=(2x2−1)=2x

Thus, the derivative is h′(x)=2x.

Substitute f′(x2) for g′(h(x)) and 2x for h′(x) in equation (3),

g′(x)=f′(x2)⋅2x=2x⋅f′(x2)

Substitute x=2 in the above equation,

g′(2)=2(2)⋅f′(22)

=4⋅f′(4) (4)

Obtain the value of f′(4).

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

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

f′(4)=f(4.1)−f(4)4.1−4

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

f′(4)=2.2−24.1−4=0.20.1=2

Thus, the value f′(4) is f′(4)=2.

Substitute 2 for f′(4) in equation (4),

g′(2)≈4⋅2=8

Therefore, the value g′(2) is approximately 8.