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Mathematics

MAT237Y1

Dan Dolderman

Fall

Description

Calc 1
Wed 10/1/08
4.1 - Implicit Diﬀerentiation
- Exam 2 in two weeks!
In case you’ve been on top of the stuﬀ on the website, you might notice that today was scheduled
as related rates. We’re just going to skip that. Today we are going to do 4.1: Implicit
Diﬀerentiation. For some reason, the book decided it would be a good idea to do related rates
ﬁrst, which is just stupid. So, this means today we will be going over 4.1, tomorrow, you’ll be
brieﬂy introduced to related rates with your worksheet, and then on Friday and Monday, we shall
work on problem after problem of related rates questions. Your homework will, of course, be
aﬀected, and so the homework assigned for tonight will be due next tuesday.
Okay, up until now, we’ve basically dealt with functions where we have y = f(x). y is on one side
of the equation and a whole bunch of x’s are on the other. No mixing and matching involved. Well,
this isn’t a perfect world, and sometimes we have both x’s and y’s on the same side. For example,
2
xy − x y + 1 = x
Well, sometimes we can actually turn that equation into a function y = f(x). In this case, we have
2
xy − x y = x − 1
so factor out a y on the left and we get
2
y(x − x ) = x − 1
or
x − 1
y =
x − x2
Okay, that’s pretty sweet. The problem is, we can’t always solve explicitely for y. Or, perhaps we
can, but it’s just so diﬃcult or perhaps once we do solve for y it’s so diﬃcult to take the
derivative, that it’s just not worth it!
Sometimes when we solve for y, we have to actually split it up into two diﬀerent functions in order
to pass the vertical line test. For example, consider the circle with center at the origin and radius
3:
x + y = 9
If we were to solve for y, we split it into two functions
p
y = 9 − x2
p
y = − 9 − x2
The ﬁrst graph is the upper semicircle, while the second is the lower semicircle. Together, they
form the entire circle (ie, just graphing both functions on the same axes).
So, a function f is deﬁned implicitly by some equation in x and y if the graph y = f(x) coincides
with a portion of the graph of the equation.
1 Example: let x = y . Note this is not a function because it fails the vertical line test! But, if we
√ √
solve for y, we get y = x and y = − x, and thinking of those two as separate functions, they
each pass the vertical line test individually. So, we say x = y implicitly deﬁnes the functions
√ √
f1(x) = x and f 2x) = − x.
So, why are we bothering to think about this stuﬀ if we can just solve for y? Well, sometimes you
can’t. For example, it is impossible to solve for y in this equation:
x + y = 3xy
Go ahead and try it! I doubt you’ll get very far... The graph of this equation is called the Folium
of Descartes, by the way, and its graph looks like this:
1
x
–2 –1 1 2 3
0
–1
y –2
–3
–4
Okay, so now let’s move on to diﬀerentiating these implicit functions. What exactly do we do?
Well, you basically just diﬀerentiate as you normally would, just keep in mind that y is a function
of x!
Example: ﬁnd the derivative of y + xy = 1.
Solution: So, there are two diﬀerent ways we can do this problem. One way is to solve for y and
then take the derivative as normal. This is how it works out:
y(x + 1) = 1
1
y =
x + 1
dy = −1
dx (x + 1)2
By the way, you shouldn’t have to use quotient rule here! You certainly can, but it’s a bit more
work than necessary. Just use the chain rule with y = (x + 1) −1. If there’s just a constant in the
numerator or in the denominator, you don’t need the quotient rule! It’s just a waste of time
because the derivative of that constant will turn out to be zero!
Anywho, here’s how to go about solving this implicitly. Just go across and diﬀerentiate everything
with respect to x. Also keep in mind that y is a function of x (i guess you can think of it as f(x) if
2 you’d like to), so if you’re diﬀerentiating xy, you need to use the product rule!
y + xy = 1
dy + d [x]y + xdy = d [1]
dx dx dx dx
dy + y + xdy = 0
dx dx
dy
And from here, just solve

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