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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

*x ^{2}y^{2}*+

> 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

Second derivatives can also be taken 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 (1,-1) 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

> with(plots):Here is an example of using this command to plot the hyperbola

> 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

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

*y ^{2}* (

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

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

> evalf(subs({x=2,y=y_sol},implicitdiff(g,y,x )));

- 1.
- Find the slope of the graph of
*x*-^{2}*y*=3 at the point (2,-1). Supply a plot of the graph that includes the point in question.^{2} - 2.
- For the relation
*x*+^{2}y^{2}*y*=0, the^{3}`implicitdiff`command gave the result Explain why this formula can't be used at the point (0,0) in terms of the graph of this relation. Does it give the correct results if you substitute*y*=0 or*y*=-*x*?^{2} - 3.
- Consider the relation
*x*+^{2}y^{2}*y*= 1.^{3}- (a)
- Use the
`implicitplot`command and/or any other technique to determine what the graph of this relation looks like. How many separate branches does the graph contain? - (b)
- Find at the point(s) on the graph
where
*x*=1. - (c)
- Are there any points on the graph where the derivative is undefined because the tangent line is vertical? If so, find them. (Hint - if the tangent line is vertical, what is true about ?)

11/30/1999