# The derivative of P ( x ) at x = 2 . ### 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 3.2, Problem 46E

(a)

To determine

## To find: The derivative of P(x) at x=2.

Expert Solution

The derivative of P(x) at x=2 is P(2)=32.

### Explanation of Solution

Given:

The function is P(x)=F(x)G(x).

Derivative rule: Product Rule

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

ddx[f1(x)f2(x)]=f1(x)ddx[f2(x)]+f2(x)ddx[f1(x)]

Calculation:

Obtain the derivative of P(x) at x=2.

P(x)=ddx(F(x)G(x))

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

P(x)=F(x)ddx(G(x))+G(x)ddx(F(x))=F(x)G(x)+G(x)F(x)

Substitute 2 for x in P(x),

P(x)=F(x)G(x)+G(x)F(x) (1)

From the given graph, it is observed that F(2)=3 and G(2)=2, the function F(x) is horizontal tangent at x=2 and the function G(x) is linear that passes through the points (0,1) and (4,3).

Obtain the values F(2) and G(2).

The slope of tangent to F(x) at x=2 is zero.

Therefore, the value F(2)=0.

The slope of the function G(x) passing through the points (0,1) and (4,3) is,

m1=3140=24=12

Therefore, the value of G(2)=12.

Substitute the values F(2)=3 , G(2)=2,F(2)=0 and G(2)=12 in equation (1),

P(2)=(3×12)+(2×0)=32+0=32

Therefore, the derivative of P(x) at x=2 is P(2)=32.

(b)

To determine

### To find: The derivative of Q(x) at x=7.

Expert Solution

The derivative of Q(x) at x=7 is Q(7)=4312.

### Explanation of Solution

Given:

The function is Q(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 derivative of Q(x) at x=7.

Q(x)=ddx(F(x)G(x))

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

Q(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 7 for x in Q(x),

Q(7)=G(7)F(7)F(7)G(7)(G(7))2 (2)

From the given graph, it is observed that F(7)=5 , G(7)=1 , the function F(x) is linear that passes through the points (3,4) and (7,5) and the function G(x) is linear that passes through the points (4,3) and (7,1).

Obtain the values F(7) and G(5).

Since the slope of the line at every point is equal, the slope of the function f(x) passing through the points (3,4) and (7,5) is,

m2=5473=14

Therefore, the value of F(7)=14.

The slope of the function G(x) passing through the points (4,3) and (7,1) is,

m3=1374=23

Therefore, the value of G(7)=23.

Substitute the values F(7)=5 , G(7)=1,F(7)=14 and G(7)=23 in equation (2),

Q(7)=(1×14)(5×23)12=14+103=312+4012=4312

Therefore, the derivative of Q(x) at x=7 is Q(7)=4312.

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