Chapter 3 | Derivatives
249
( x + h ) 3 − x 3 h
⎛ ⎝ x 3
⎞ ⎠ = lim h →0
d dx
Notice that the first term in the expansion of ( x + h ) 3 is x 3 and the second term is3 x 2 h .All other terms contain powers of h that are two or greater. In this step the x 3 terms have been cancelled, leaving only terms containing h .
x 3 +3 x 2 h +3 xh 2 + h 3 − x 3 h
= lim
h →0
3 x 2 h +3 xh 2 + h 3 h h (3 x 2 +3 xh + h 2 ) h (3 x 2 +3 xh + h 2 )
= lim
h →0
= lim
Factor out the common factor of h .
h →0
After cancelling the common factor of h , the only term not containing h is3 x 2 .
= lim
h →0
=3 x 2
Let h go to 0.
⎛ ⎝ x 4
⎞ ⎠ .
Find d dx
3.12
As we shall see, the procedure for finding the derivative of the general form f ( x ) = x n is very similar. Although it is often unwise to draw general conclusions from specific examples, we note that when we differentiate f ( x ) = x 3 , the power on x becomes the coefficient of x 2 in the derivative and the power on x in the derivative decreases by 1. The following theorem states that the power rule holds for all positive integer powers of x . We will eventually extend this result to negative integer powers. Later, we will see that this rule may also be extended first to rational powers of x and then to arbitrary powers of x . Be aware, however, that this rule does not apply to functions in which a constant is raised to a variable power, such as f ( x ) =3 x .
Theorem 3.3: The Power Rule Let n be a positive integer. If f ( x ) = x n , then
f ′( x ) = nx n −1 .
Alternatively, we may express this rule as
d dx
x n = nx n −1 .
Proof For f ( x ) = x n where n is a positive integer, we have
( x + h ) n − x n h .
f ′( x ) = lim h →0
⎛ ⎝
⎞ ⎠ x n −3 h 3 +…+ nxh n −1 + h n ,
Since ( x + h ) n = x n + nx n −1 h + ⎛ ⎝
⎞ ⎠ x n −2 h 2 +
n 3
n 2
we see that
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