So far, the majority of the material discussed in class was on the following
type of problem: Given an equation of the form
y = f (x),
compute
dy
dx .
Example 0.1. If y = x3 + 5x2 , then we have that
dy
= 3x2 + 10x.
dx
But what happens if the relationship between x and y is more complicated?
What if it’s annoying, difficult, or impossible to get an equation relating x and
y in the form y = f (x)?
Problem 0.2. Compute
dy
dx
if
x2 + 5xy − y 2 + 2x − 3y = 1.
Before starting, it’s good to get a mental map of what kind of problem this
is.
What kind of graph does this equation describe? This has the form of a conic
section, and using the fact that the discriminant is positive, we can conclude
that this particular conic section is a hyperbola.
dy
There are two ways we can go about the problem of finding dx
: one way is
to write y explicitly as a function of x. In this example, it’s more workable than
it may seem at first- you can do this by applying the quadratic formula in the
y-variable. This is a huge pain, so we would rather avoid doing so.
Instead of “explicitly” solving for y and then differentiating, we prefer to
directly differentiate both sides of the “implicit” equation relating x and y.
d 2
x + 5xy − y 2 + 2x − 3y = 0
dx
As usual, the derivative of the sum is the sum of the derivatives, and we can
pull out constants:
d 2
d
d 2
d
d
x + 5 xy −
y +2 x−3 y =0
dx
dx
dx
dx
dx
The terms that don’t contain a y are easy- we just differentiate them normally.
For the terms that do contain a y, we use the product and chain rules together
dy
d
with the fact that dx
y = dx
:
2x + 5y + 5x
Now, we solve for
dy
dy
dy
− 2y
+2−3
=0
dx
dx
dx
dy
dx :
(5x − 2y − 3)
dy
= −2x − 5y − 2
dx
dy
−2x − 5y − 2
=
dx
5x − 2y − 3
1
The downside to this kind of answer is that it depends on both x and y- we call
this type of equation a non-autonomous differential equation.
However, in practice, this is often not a problem. For example, let’s consider
the following problem:
Problem 0.3. Find the tangent lines to the graph of
x2 + 5xy − y 2 + 2x − 3y = 1.
at the points for which x = 1.
Looking at the graph of the hyperbola in the geogebra example file, it’s clear
that there are two points on the hyperbola for which x = 1. We can find both
of these points using algebra: substituting x = 1 into the above equation gives
1 + 5y − y 2 + 2 − 3y = 1
which simplifies to
−y 2 + 2y + 2 = 0
I like to keep the leading coefficient positive:
y 2 − 2y − 2 = 0
We can now solve for y using the quadratic formula:
√
4+8
2√
2±2 3
y=
2√
y =1± 3
y=
2±
dy
, we just need to plug in the appropriate x and y
So to find the slope dx
values into the expression we calculated before:
−2x − 5y − 2
dy
=
dx
5x − 2y − 3
First, we plug in (1, 1 +
√
3) to get
√
dy
−2 − 5(1 + 3) − 2
√
=
dx
5 − 2(1 + 3) − 3
which comes out to
√
−9 − 5 3
√
−2 3
√
√
which is equal to 3 2 3 + 52 . A similar calculation with y = 1 − 3 gives
for the slope. Therefore, we have the two equations for the lines:
!
√
√
3 3 5
y − (1 + 3) =
(x − 1)
+
2
2
2
√
−3 3
2
+ 52
√
y − (1 − 3) =
!
√
−3 3 5
(x − 1).
+
2
2
These two tangent lines are graphed in the geogebra file.
I don’t expect to reach this point in the lecture notes, but I’ll put it in
anyway.
dy
We’ll do another example for practice: Find dx
at the point (1, 1) on the
graph of
x2 y − 2xy 2 = −1.
(You can check that (1, 1) is on the graph). We differentiate both sides to get
dy
dy
= 0.
− 2 y 2 + 2xy
2xy + x2
dx
dx
we put the
dy
dx ’s
together:
(x2 − 4xy)
and divide:
and plug in (3, 1) :
which comes out to 0.
dy
= 2y 2 − 2xy
dx
dy
2y 2 − 2xy
= 2
dx
x − 4xy
0
dy
=
dx
9−4
3