# The derivative of u ( x ) at x = 1 .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.4, Problem 55E

(a)

To determine

## To find: The derivative of u(x) at x=1.

Expert Solution

The derivative of u(x) at x=1 is u(1)=34.

### Explanation of Solution

Given:

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

Result used: Chain Rule

If h is differentiable at x and g is differentiable at h(x), then the composite function F=gh 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 y2y1x2x1 .

Calculation:

Obtain the value of u(1).

u(x)=ddx(f(g(x)))

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

ddx(u(x))=ddx(f(g(x)))u(x)=f(g(x))g(x)

Substitute x=1 in the above equation,

u(1)=f(g(1))g(1)

From the given graph, it is observed that g(1)=3.

u(1)=f(3)g(1) (2)

From the given graph, the function f(x) is linear from (2,4) to (6,3) and the function g(x) is linear that passes through the points (0,6) and (2,0).

Obtain the values f(3) and g(1).

Since the slope of the line at every point is equal, the slope of the function f(x) passes through the points (2,4) and (6,3) is,

m1=3462=14

Therefore, the value of f(3)=14.

The slope of the function g(x) passing through the points (0,6) and (2,0) is,

m2=0620=62=3

Therefore, the value of g(1)=3.

Substitute 14 for f(3) and 3 for g(1) in equation (2),

u(1)=(14)(3)=34

Therefore, the value u(1) is u(1)=34.

(b)

To determine

### To find: The value v′(1).

Expert Solution

The function v(1) does not exist.

### Explanation of Solution

Given:

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

Calculation:

Obtain the derivative of v(x)=g(f(x)).

v(x)=ddx(v(x))=ddx(g(f(x)))

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

v(x)=g(f(x))f(x)

Substitute x=1 in the above equation,

v(1)=g(f(1))f(1)

From the graph observe that,f(1)=2.

v(1)=g(2)f(1) (3)

Obtain the slope of g(2).

From the given graph, it is observed that the function g(x) has infinitely many tangent lines at x=2. That is, g(2) does not exist.

Therefore, it can be concluded that the function v(1) does not exist.

(c)

To determine

### To find: The value w′(1).

Expert Solution

The value w(1) is w(1)=34.

### Explanation of Solution

Given:

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

Calculation:

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

w(x)=ddx(w(x))=ddx(g(g(x)))

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

w(x)=g(g(x))g(x)

Substitute x=1 in the above equation,

w(1)=g(g(1))g(1)

From the graph observe that,g(1)=3.

u(1)=g(3)g(1) (4)

The function g(x) is linear that passes through the points (2,0) and (5,2) and the function g(x) is linear that passes through the points (0,6) and (2,0).

Obtain the value g(3) and g(1).

Since the slope of the line at every point is equal, the slope of the function g(x) passes through the points (2,0) and (5,2) is,

m3=2052=23

Therefore, the value of g(3)=23.

Obtain the value g(1).

The slope of the function g(x) passing through the points (0,6) and (2,0) is,

m4=0620=62=3

Therefore, the value of g(1)=3.

Substitute 23 for g(3) and 3 for g(1) in equation (2),

u(1)=(23)(3)=2

Therefore, the derivative of w(1) is w(1)=2.

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