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 32E
To determine

To find: The derivative of the function f(x)=41x at x=a.

Expert Solution

Answer to Problem 32E

The derivative of the function f(x) at x=a is 2(1a)32_.

Explanation of Solution

Formula used:

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

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

Difference of square formula: (a2b2)=(a+b)(ab).

Calculation:

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

Compute f(a) by using the equation (1).

f(a)=limh0f(a+h)f(a)h=limh041(a+h)41ah=limh041a41(a+h)1a1(a+h)h=limh04(1a1(a+h))h1a1(a+h)

Multiply both the numerator and the denominator by the conjugate of the numerator,

f(a)=limh04(1a1(a+h))h1a1(a+h)×(1a+1(a+h))(1a+1(a+h))

Apply the difference of squares formula,

f(a)=limh04((1a)2(1(a+h))2)h1a1(a+h)(1a+1(a+h))=limh04(1a(1(a+h)))h((1a)1(a+h)+(1(a+h))1a)=limh04((1a)(1ah))h((1a)1(a+h)+(1(a+h))1a)=limh04hh((1a)1(a+h)+(1(a+h))1a)

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)=limh04((1a)1(a+h)+(1(a+h))1a)=4((1a)1(a+0)+(1(a+0))1a)=4((1a)1a+(1a)1a)=42(1a)1a

=2(1a)32

Thus, the derivative of the function f(x) at x=a is 2(1a)32_.

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