If we need to calculate the derivative of a function at a point, there are two ways we can think about doing this. For example suppose f(x) = x2 - z, and we need to determine the value of f(4).
If we need to calculate the derivative of a function at a point, there are two ways we can think about doing this. For example suppose f(x) = x2 - z, and we need to determine the value of f(4).
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter2: Functions
Section: Chapter Questions
Problem 7T
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![If we need to calculate the derivative of a function at a point, there are two ways we can think about doing
this. For example suppose f() = x – x, and we need to determine the value of f(4).
One option is to just calculate the derivative at that point by plugging the point into the limit definition, like
this:
f(4 + h) – f(4)
(
(4 + h)² – (4 + h)) - (4² – 4)
f'(4) = lim
(16 + 8h + h2 -4 – h) - (16 - 4)
lim
7h + h2
= lim
h(7+ h)
lim
h
= lim
h→0
h
lim 7+h = 7.
||
h
h
h
Another option is to calculate f'(x) in terms of x, and then plug in a value for x at the end, like this:
f(x + h) – f(x)
(2+ h) – (x + h)) - (2² – a)
(22 + 2xh + h² – z – h) – (22 – 2)
lim
f'(x) = lim
= lim
2xh + h2 - h
h(2x + h - 1)
= lim
h 0
h
h
= lim
= lim
h
h
h
that is to say, f'(x)
= 2x – 1, and therefore f'(4) = 2(4) – 1 = 7.
Notice that we get the answer f'(4) = 7 both ways. The second approach might look somewhat more
complicated at first, but it turns out to be much more efficient if we ever need to know the value of the
derivative at multiple points. For example, if we now wanted to know f' (1), f'(7) and f'(11), we wouldn't
need to calculate the limit definitions all over again
formula f'(x) = 2x – 1.
-- we could just plug the numbers 1, 7 and 11 into the
Let g(x) = x – x².
g(3 + h) – g(3)
Find g'(3) by calculating lim
h-0
h
g(x + h) – g(x)
Next, find g'(æ) by calculating lim
h 0
h
Finally, find g'(3) by plugging 3 into g'(x).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb6fa906-3c7d-40af-a4e1-bd5914abe0ff%2Fc8954931-7f1f-417d-bb91-dd71949818fd%2Fjymyx6n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:If we need to calculate the derivative of a function at a point, there are two ways we can think about doing
this. For example suppose f() = x – x, and we need to determine the value of f(4).
One option is to just calculate the derivative at that point by plugging the point into the limit definition, like
this:
f(4 + h) – f(4)
(
(4 + h)² – (4 + h)) - (4² – 4)
f'(4) = lim
(16 + 8h + h2 -4 – h) - (16 - 4)
lim
7h + h2
= lim
h(7+ h)
lim
h
= lim
h→0
h
lim 7+h = 7.
||
h
h
h
Another option is to calculate f'(x) in terms of x, and then plug in a value for x at the end, like this:
f(x + h) – f(x)
(2+ h) – (x + h)) - (2² – a)
(22 + 2xh + h² – z – h) – (22 – 2)
lim
f'(x) = lim
= lim
2xh + h2 - h
h(2x + h - 1)
= lim
h 0
h
h
= lim
= lim
h
h
h
that is to say, f'(x)
= 2x – 1, and therefore f'(4) = 2(4) – 1 = 7.
Notice that we get the answer f'(4) = 7 both ways. The second approach might look somewhat more
complicated at first, but it turns out to be much more efficient if we ever need to know the value of the
derivative at multiple points. For example, if we now wanted to know f' (1), f'(7) and f'(11), we wouldn't
need to calculate the limit definitions all over again
formula f'(x) = 2x – 1.
-- we could just plug the numbers 1, 7 and 11 into the
Let g(x) = x – x².
g(3 + h) – g(3)
Find g'(3) by calculating lim
h-0
h
g(x + h) – g(x)
Next, find g'(æ) by calculating lim
h 0
h
Finally, find g'(3) by plugging 3 into g'(x).
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