The **implicitdiff** command can be used to find derivatives of
implicitly defined functions. Suppose we wanted to use implicit
differentiation to find
for the relation

Then we first define our relation and give it a label for later use.

> f:=x^2*y^2+y^3=0;The syntax of the

> implicitdiff(f,y,x);

The result of the command is the implicit derivative,
. The syntax of this command is very similar to that of
the `diff` command. The first argument is always the relation
that you want to differentiate implicitly. We were careful to use an
equation for this argument, but if you just give an expression for
this argument, Maple assumes you want to set this expression equal to
zero before differentiating. The second argument to the
`implicitdiff` command is where you tell Maple what the
dependent variable is. That is, by putting `y` here, we were
saying that we were thinking of this relation as defining and
not . The remaining arguments to `implicitdiff` are for
specifying the order of the derivative you want.

Second derivatives can also be computed with **implicitdiff**. The
following command computes
.

> implicitdiff(f,y,x,x);

To compute numerical values of derivatives obtained by implicit differentiation, you have to use the subs command. For example, to find the value of at the point you could use the following command.

> subs({x=1,y=-1},implicitdiff(f,y,x));

Sometimes you want the value of a derivative, but first have to find
the coordinates of the point. More than likely, you will have to use
the `fsolve` command for this. However, to get the
`fsolve` command to give you the solution you want, you often
have to specify a range for the variable. Being able to plot the graph
of a relation can be a big help in this task, so we now describe the
`implicitplot` command.
This Maple command for plotting implicitly defined functions
is in the `plots` package which must be loaded before using the
command.

> with(plots):Here is an example of using this command to plot the hyperbola . Note that you have to specify both an range and a range. This is because the

> implicitplot(x^2-y^2=1,x=-3..3,y=-3..3);To get a good graph with this command, you usually have to experiment with the ranges. For example the following command

> implicitplot(f,x=-1..1,y=1..2);produces an empty plot. The reason is simply that there are no solutions to with . This is easy to see if you rewrite the equation as and recognize that both sides of the equation must be nonnegative. Usually a good strategy to follow is to start with fairly large ranges, for example to for both variables, and then refine them based on what you see.

This command can also have problems if the relation in question has solution branches that cross or are too close together. For example, try the following command.

> implicitplot(f,x=-1..1,y=-1..0);For less than about , you should see the two smooth curves. However, for values of closer to zero the two curves become jagged. To understand this, we need to take a closer look at the relation we tried to plot. The key is to notice that we can factor out and write our relation as follows.

This makes it clear that the graph of the relation really has two pieces: and . These two curves intersect at the origin, which explains why

As our last example, consider the relation . Try the following commands to see what a part of the graph of this relation looks like.

> g := x^2*sin(y)=1; > implicitplot(g,x=-4..4,y=-10..10);Suppose you were asked to find the slope of the graph of this relation at , but you were only given that the value of was about 9. Using the plot, it is relatively easy to find this derivative by first using

> y_sol := fsolve(subs(x=2,g),y=8..10); > evalf(subs({x=2,y=y_sol},implicitdiff(g,y,x)));

- Find the slope of the graph of at the point
. Supply a plot of the relation and the tangent line at the point in
question.
- For the relation from the first exercise, find
the second derivative
at the point
.
- Consider the relation
.
- Use the
`implicitplot`command to plot the graph of this relation for the region and . - Find at the point on the graph where and the value is negative.

- Use the

2001-10-09