Calculus Volume 1

256

Chapter 3 | Derivatives

Simplifying, we obtain

2 +30 x

k ′( x ) = 20 x

.

(4 x +3) 2

Find the derivative of h ( x ) = 3 x +1 4 x −3 .

3.17

It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form x k where k is a negative integer.

Theorem 3.7: Extended Power Rule If k is a negative integer, then

⎛ ⎝ x k

⎞ ⎠ = kx k −1 .

d dx

Proof If k is a negative integer, we may set n =− k , so that n is a positive integer with k =− n . Since for each positive integer n , x − n = 1 x n , we may now apply the quotient rule by setting f ( x ) =1 and g ( x ) = x n . In this case, f ′( x ) =0 and g ′( x ) = nx n −1 . Thus,

0( x n )−1 ⎛

⎞ ⎠

⎝ nx n −1

d dx

( x − n ) =

.

( x n ) 2

Simplifying, we see that

n −1

d dx

( x − n ) = − nx

=− nx ( n −1)−2 n =− nx − n −1 .

x 2 n

Finally, observe that since k =− n , by substituting we have d dx ⎛ ⎝ x k ⎞

⎠ = kx k −1 .

Example 3.26 Using the Extended Power Rule

⎛ ⎝ x −4

⎞ ⎠ .

Find d dx

Solution By applying the extended power rule with k =−4, we obtain

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