false

Class Notes
(836,321)

Canada
(509,732)

University of Toronto St. George
(44,025)

Mathematics
(2,843)

MAT237Y1
(53)

Dan Dolderman
(31)

Lecture

Unlock Document

Mathematics

MAT237Y1

Dan Dolderman

Fall

Description

Calc 1
Fri 9/26/08
3.4 - Product and Quotient Rules
Okay, so I mentioned earlier that derivative of the product of two functions is not the same thing
2
as the product of the derivatives. Here’s why not: let f(x) ▯ x a▯▯ g(x) =▯x. The derivative of
d 3 2 d d 2
the product is x = 3x , but the product of the derivatives is x x = 2x. The last
dx dx dx
time I checked, those are not the same (well, they are if x = 0, but you know what I mean). So,
what exactly happens when you take the derivative of the product? Well, let’s see, using (what
else but) the de▯nition of the derivative!
Product Rule:
d f(x + h)g(x + h) ▯ f(x)g(x)
[f(x)g(x)] = lim
dx h!0 h
f(x + h)g(x + h) + f(x + h)g(x) ▯ f(x + h)g(x) ▯ f(x)g(x)
= lim
h!0 h
Whoa, what just happened there? We added zero! This is something typical that happens where
after you know what to do, it’s pretty obvious that it’s correct; the hard part is knowing that
you’re supposed to \add zero" there! Let’s continue so you can see why I just added and
subtracted that f(x + h)g(x):
▯ ▯
f(x + h)g(x + h) ▯ f(x + h)g(x) f(x + h)g(x) ▯ f(x)g(x)
= lim +
h!0▯ h ▯
g(x + h) ▯ g(x) f(x + h) ▯ f(x)
= lim f(x + h) + g(x)
h!0 h h
▯ ▯▯ ▯ ▯ ▯▯ ▯
= lim f(x + h) lim g(x + h) ▯ g(x) + lim g(x) limf(x + h) ▯ f(x)
h!0 h!0 h h!0 h!0 h
d d
= f(x) ▯ dx [g(x)] + g(x)dx [f(x)]
0 0 0
So, in other words, (fg) = fg + gf . Make sure you memorize this! You WILL be tested on it!
Example: ▯nd dy if y = (x + 2)(5x ▯ 4x ▯ 1).
dx
0 3 d 4 2 4 2 d 3
y = (x + 2) (5x ▯ 4x ▯ 1) + (5x ▯ 4x ▯ 1) (x + 2)
3 dx 3 4 2 2 dx
= (x + 2)(20x ▯ 8x) + (5x ▯ 4x ▯ 1)(3x )
This is totally correct, and you can leave your answer just like this. However, I’m going to expand
it out so that we can compare.
1 3 3 4 2 2 6 4 3 6 4 2
(x + 2)(20x ▯ 8x) + (5x ▯ 4x ▯ 1)(3x ) = 20x ▯ 8x + 40x ▯ 16x + 15x ▯ 12x ▯ 3x
6 4 3 2
= 35x ▯ 20x + 40x ▯ 3x ▯ 16x
So, let’s now expand in the beginning and then take the derivative and see if we get the same
answer (something you do if you forget the product rule. However, be careful because on this test,
we may ask you to speci▯cally to take the derivative using the product rule, so if you don’t
remember, you’re screwed!).
7 5 3 4 2
y = 5x ▯ 4x ▯ x + 10x ▯ 8x ▯ 2
0 6 4 3 2
y = 35x ▯ 20x + 40x ▯ 3x ▯ 16x
Well, by golly! We got the same answer! Both ways are completely correct; however, many times,
this second method we just used is much more tedious and usually involves a lot more busywork.
Your best method is to just memorize the product rule!
0 p 3=2 2
Example: ▯nd f (x) where f(x) = x(x ▯ x + 6x).
p d d p
f (x) = x (x3=2▯ x + 6x) + (x3=2▯ x + 6x) x
dx dx
p ▯ p ▯
= x 3 x ▯ 2x + 6 + (x3=2▯ x + 6x) p1
2 2 x
Again, this is totally correct. You don’t have to simplify if it doesn’t ask you to. If you would like,
you can go ahead and expand ▯rst and then take the derivative, like in the previous example and
see if you get the same thing.
Okay, just like how we have problems with the product of two functions when we take the
derivative, we also have problems when we divide them. So, let’s derive the quotient rule!
Quotient Rule:
▯ ▯ f(x+h) f(x)
d f(x) g(x+h)▯ g(x)
= lim
dx g(x) h!0 h
f(x+h)g(x) f(x)g(x+h)
g(x+h)g(x) g(x)g(x+h)
= lim
h!0 h
f(x+h)g(x)▯f(x)g(x+h)
g(x+h)g(x)
= lim
h!0 h
f(x + h)g(x) ▯ f(x)g(x + h) + f(x

More
Less
Related notes for MAT237Y1

Join OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.