This document discusses implicit differentiation, which is a technique for taking the derivative of equations that cannot be solved explicitly for y as a function of x. It explains that when differentiating terms involving both x and y, the derivative of the y term is dy/dx. As an example, it shows the differentiation of xy using the product rule, which yields y + x*dy/dx. The document concludes by applying this technique to differentiate the equation y4 + xy = x3 - x + 2 implicitly with respect to x.

1. Implicit Dierentiation
Tyler Murphy
October 8, 2014
How do we deal with equations which may not be functions? That is, a lot of equations
we have dealt with so far are easily put into the form y = stuff. However, some equations
are not so easily turned into that form. Or they may result in many dierent functions for
y. For example, x2 + y2 = 1 can be solved for y into y =
p
1 x2, which is two separate
functions of y. But consider,
y4 + xy = x3 x + 2
This equation can't be easily solved for y.
This is where implicit dierentiation comes in play. Here's the mechanics of how it works.
If you are taking the derivative with respect to a certain variable (usually x or t), then
each term you dierentiate with a dierent variable, you take the derivative like normal
and then add a dy
dx at the end of the term. (y is the variable not the same as the one are
dierentiating with respect to - in this case, x)
For example,
d
dx
(xy)
This is an instance of the product rule. x y. First, we make a list of our functions f
and g and their derivatives.
f = x
f0 = 1
g = y
g0 = 1 dy
dx
Putting these into our product rule (f0g + g0f), we get:
(1)(y) + x(1)
dy
dx
= y + x
dy
dx
If we were solving an equation instead of just evaluating the expression, we would next
solve for dy
dx .
1

2. Now let's use this information to

3. nd the derivative of
y4 + xy = x3 x + 2
First we take the derivative with each term, adding a dy
dx after each term involving y.
Note that we have already found the derivative of xy above.
) 4y3 dy
dx
+ (y + x
dy
dx
) = 3x2 1
)
dy
dx
(4y3 + x) = 3x2 y 1
)
dy
dx
=
3x2 y 1
4y3 + x
2