> f:=x->x^2+7*x+5; > g:=x^3-5*x+8;Here are some examples that show how

> D(f)(x); > diff(g,x); > diff(f(x),x);See what happens when the function or expression notations are used incorrectly.

> diff(g(x),x); > diff(f,x);After the last four examples, you should be convinced that proper notation is very important in doing derivatives in Maple. Maple can also do higher derivatives. Check these commands. Again, pay attention to the difference between expressions and functions.

> diff(g,x,x); > diff(g,x,x,x); > (D@@2)(f)(x); > (D@@3)(f)(x);If you want to evaluate the higher derivative at a specific value of

> (D@@2)(f)(2); > subs(x=3,diff(g,x));Suppose you wanted to find the equation of the tangent line to the graph of

> tanline := D(f)(5)*(x-5)+f(5);More information on

> f:=x^2*y^2+y^3=0; > implicitdiff(f,y,x);where

> g:=x^2+y^3=1; > implicitdiff(g,y,x);Second derivatives can also be taken with

> implicitdiff(g,y,x,x);Maple also has a command for plotting implicitly defined functions. It is in the package

> with(plots): > implicitplot(x^2-y^2=1,x=-3..3,y=-3..3);Suppose you want to find the equation of the line tangent to the graph of defined implicitly. For instance, find the equation of the line tangent to at the point . You can use the point-slope form of a line in implicit form to get the equation of the tangent line. The Maple commands below show how this can be done.

> f:=x^2*y^2+y^3=0; > m:=subs({x=1,y=-1},implicitdiff(f,y,x)); > tanline := y-(-1)=m*(x-1); > implicitplot({f,tanline},x=0..2,y=-3..0);

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 `solve` or `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. Plotting the graph of a relation can be a big help in this task. For instance, if you wanted the slope of as in the previous example at and is negative, but the value is not given, then you would first need to solve for by substituting the value into and then solve for . See how this is done below.

> f:=x^2*y^2+y^3=0; > solve(subs(x=1,f),y);

- Given
, evaluate the second derivative at
using function notation and
- the
**D**command. - the
**diff**command.

- the
- Find the equation of the line tangent to the graph of
at the point where and is negative. Include a plot of the relation and the tangent line on the same graph over the interval
and
.
- Consider the graph defined implicitly by the equation
.
- Enter the equation, calling it
*h*. - Use the
**implicitplot**command to verify visually that the graph is a hyperbola. - Find the slopes to this curve at the two points where it intersects the x-axis labeling them
*m1*and*m2*. (Hint: You will first need to use the**solve**command to find the y-values) What can you tell about the tangent lines at these points given your answers for the slopes? - Find the equations of the tangent lines to this curve at the two points where it intersects the x-axis labeling them
*t1*and*t2*. Remenber to enter the equations implicitly. - Graph the ellipse and the two tangent lines on one graph.

- Enter the equation, calling it

2006-01-31