BuyFind

Single Variable Calculus: Concepts...

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

Single Variable Calculus: Concepts...

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

Solutions

Chapter 3.2, Problem 45E

(a)

To determine

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

Expert Solution

Answer to Problem 45E

The derivative of u(x) at x=1 is u(1)=0.

Explanation of Solution

Given:

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

Derivative rule:

Product Rule: ddx[f1(x)f2(x)]=f1(x)ddx[f2(x)]+f2(x)ddx[f1(x)]

Calculation:

Obtain the value of u(1).

u(x)=ddx(f(x)g(x)) 

Apply the product rule (1) and simplify the terms,

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

Substitute 1 for x in u(x),

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

From the given graph, it is observed that f(1)=2 , g(1)=1 , the function f(x) is linear from (0,0) to (2,4) and the function g(x) is linear that passes through the points (0,2) and (2,0).

Obtain the values f(1) 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 (0,0) and (2,4) is,

m1=4020=2

Therefore, the value of f(1)=2.

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

m2=0220=1

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

Substitute the values f(1)=2,g(1)=1,f(1)=2 and g(1)=1 in equation (1),

u(1)=(2×1)+(1×2)=2+2=0

Therefore, the derivative of u(x) at x=1 is u(1)=0.

(b)

To determine

To find: The derivative of v(x) at x=5.

Expert Solution

Answer to Problem 45E

The derivative of v(x) at x=5 is v(5)=23.

Explanation of Solution

Given:

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

Derivative rule:

Quotient Rule: If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]f1(x)ddx[f2(x)][f2x]2

Calculation:

Obtain the value of v(5).

v(x)=ddx(f(x)g(x)) 

Apply the Quotient rule (1) and simplify the terms,

v(x)=g(x)ddx(f(x))f(x)ddx(g(x))(g(x))2=g(x)f(x)f(x)g(x)(g(x))2

Substitute 5 for x in v(x),

v(5)=g(5)f(5)f(5)g(5)(g(5))2 (2)

From the given graph, it is observed that f(5)=3 , g(5)=2, the function f(x) is linear that passes through the points (2,4) and (5,3) and the function g(x) is linear that passes through the points (2,0) and (5,2).

Obtain the values f(5) and g(5).

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 (5,3) is,

m3=3452=13

Therefore, the value of f(5)=13.

The slope of the function g(x) passing through the points (2,0) and (5,2) is,

m3=2052=23

Therefore, the value of g(5)=23.

Substitute the values f(5)=3 , g(5)=2,f(5)=13 and g(5)=23 in equation (2),

v(5)=(2×13)(3×23)22=236322=83×4=23

Therefore, the derivative of v(x) at x=5 is v(5)=23.

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