The * Folium of Descartes* is given by the
equation . Picture:

The graph consists of all points which satisfy the equation. For example, is on the graph, because , , satisfies the equation.

Observe, however, that the graph is not the graph of a function. Some values of x give rise to multiple values for y. (Geomtrically, this means you can draw vertical lines which hit the graph more than once.)

Moreover, it would be difficult to solve the equation for y in terms of x (unless you happen to know the general cubic formula).

However, small pieces of the graph *do* look like function
graphs. You only need to be careful not to take a piece which is so
large that it violates the vertical line criterion. For such a piece,
the equation defines a function * implicitly*.
"Implicitly" means that y may not be solved for in terms of
x, but a given x still "produces" a unique y.

On such a small piece of the graph, it would make sense to ask for the derivative . Since it's difficult to solve for y, it's not clear how to compute the derivative.

The idea is to differentiate the equation *as is*, making
careful use of the Chain Rule. This produces another equation, from
which you can get (perhaps implicitly as well).

Differentiate term-by-term with respect to x. First, the derivative of with respect to x is :

The derivative of *with respect to
y* would be , but I'm differentiating
with respect to x, so I use the Chain Rule. Differentiate the outer
(cube) function, holding the inner function (y) fixed. Then
differentiate the inner function. I obtain

Finally, differentiate the right side . The 3 is constant, but is a *product*: Use the Product Rule.
Remember, however, that the derivative of the second factor (y) is
!

I can solve this equation for :

This may seem strange --- I've found in terms of y --- but I can use this expression for the derivative as I normally would.

For example, I'll find the values of x where the graph has a
horizontal tangent. As usual, set . I get , so . Plug this back into the original equation
(because I'm looking for points *on this curve*):

Therefore, or .

* Example.* Find the equation of the tangent line
to

Differentiate the equation implicitly:

Since I have a point to plug in, I don't solve this equation for . Instead, I plug the point in first, then solve for . Set and :

The tangent line is

You can see that the rule of thumb is: To differentiate a y-expression, differentiate it "like usual", but tack on a . The comes from the Chain Rule. Thus:

* Example.* Find the equation of the tangent line
to

Differentiate implicitly:

Set and and solve for :

The tangent line is

* Example.* Find the equation of the tangent line
to

Differentiate implicitly:

Set and and solve for :

Therefore, the tangent line is

* Example.* Find the equation of the tangent line
to

Differentiate implicitly:

Set and and solve for :

Therefore, the tangent line is

* Example.* Find the points on the following
curve where the tangent line is horizontal:

Differentiate implicitly:

I want the horizontal tangents, so set and solve for x:

To get the y-coordinates, plug into the original equation and solve for y:

This gives and .

The points are and .

* Example.* The * inverse
tangent* function satisfies

It is the * inverse function* to the tangent
function: roughly, a function which "undoes" the effect of
the tangent function.

Use implicit differentiation to compute the derivative of .

Start with and take the tangent of both sides: . Now differentiate implicitly:

I want to express the right side in terms of x. Now means that I have the following triangle:

Therefore, , and . Hence,

* Example.* Find at for

First, I'll differentiate implicitly and find . Then I'll differentiate implicitly a second time to find .

Differentiate implicitly:

Plug in and and solve for :

Next, differentiate (*) implicitly:

Note that I used the Product Rule to differentiate the term .

Now plug in , , and and solve for :

Copyright 2018 by Bruce Ikenaga