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 2.6, Problem 29E
To determine

To find: The derivative of the function f(t)=2t+1t+3 at t=a.

Expert Solution

Answer to Problem 29E

The derivative of the function f(t) at t=a is 5(a+3)2_.

Explanation of Solution

Formula used:

The derivative of a function at a number a, denoted by f(a) is,

f(a)=limh0f(a+h)f(a)h (1)

Calculation:

Obtain the derivative of the function f(t) at t=a.

Use the equation (1) to compute f(a).

f(a)=limh0f(a+h)f(a)h=limh0(2(a+h)+1(a+h)+3)(2a+1a+3)h=limh0(2(a+h)+1)(a+3)(2a+1)((a+h)+3)((a+h)+3)(a+3)h=limh0(2a+2h+1)(a+3)(2a+1)(a+h+3)h(a+h+3)(a+3)

Expand the numerator and simplifying further as follows,

f(h)=limh0a(2a+2h+1)+3(2a+2h+1)2a(a+h+3)1(a+h+3)h(a+h+3)(a+3)=limh02a2+2ah+a+6a+6h+32a22ah6aah3h(a+h+3)(a+3)=limh0(2a22a2)+(2ah2ah)+(aa)+(6a6a)+(6hh)+(33)h(a+h+3)(a+3)=limh05hh(a+h+3)(a+3)

Since the limit h approaches zero but not equal to zero, cancel the common term h from both the numerator and the denominator,

f(h)=limh05(a+h+3)(a+3)=5(a+(0)+3)(a+3)=5(a+3)(a+3)=5(a+3)2

Thus, the derivative of the function f(t) at t=a is 5(a+3)2_.

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