Implicit
Differentiation
Names:
:
by Tom
Linton, http://www.central.edu/homepages/lintont
Consider the plot of "all pairs", (x ,y) which satisfy
y^3 - xy = -6, along with the (dotted) line x = 7:
![[Maple Plot]](images/impdiff1.gif)
The graph indicates that there are 3 such pairs
with x = 7. Substituting x = 7 into the equation and rewriting so that
the right hand side is zero, yields y^3 - 7y + 6 = 0. Although this equation
has 3 nice solutions, in general, cubics are too difficult to solve by
hand, so we'll use the Solve features of our calculators to find
these 3 values of y. For starters, estimate the 3 values of y for which
y^3 - 7y + 6 = 0. These are the 3 values of y where the dotted line crosses
the plot above.
All of the TI-8x calculators (with x = 2 or more) have some sort of
"solver". The TI-82 has the most limited solver. It is limited to solving
equations for one variable and uses a non-interactive command approach.
The other versions are all similar to the TI-83 solver, which I'll cover
below.
TI-82 Solve Command:
The Solve command is item 0:Solve on the [MATH]
menu. To solve f(X) = 0 for X, where the solution you want
is close to X = guess and definitely between X = a and X
= b, (a must be less than b here) enter the command
solve(f(X), X, guess, {a,b})
For example, the smallest value of y which satisfies y^3 - 7y + 6 = 0 is
near y = -3 and definitely between y = -4 and -2, so the command solve(Y^3
- 7Y + 6, Y, -3, {-4,-2}) will give an approximate value for this
solution.
Using your solver, verify that the three values of Y are indeed Y =
-3, 1 and 2 (which you can also verify by hand).
Other TI Solvers:
The remaining TI calculators employ an interactive solver. The equations
you solve can be quite complex, involving several variables, but all variables
except one, must be set to some numeric value (meaning you can really only
solve one variable equations). The interactive solvers provide a screen
to enter the equation to solve, give values for all but one variable, designate
a guess at the solution value and provide bounds for where to look for
solutions. The TI-83 (which I'll explain) requires that you solve an equation
of the form: eqn = 0, so it's best to rewrite the equation so it
has this form. Start the interactive solver by selecting item 0:Solver...
on the [MATH] menu. If you've used the solver before, the equation
will be displayed on the top of the screen along with several other lines.
Otherwise, the screen will appear waiting for you to enter the equation
in the form 0 = "formula to solve". Move to the equation line and
enter 0 = Y^3 - X*Y + 6 as the equation, then press [ENTER]. Move to the
X = line and enter the value of 7 (since we want X = 7). Change the bounds
to be {-4,-2} (because we are looking for the value of Y in this range)
and then move to the Y = line and type in a guess for the value of Y which
solves this equation (use your guess from above). With the cursor on the
line for the variable you would like to solve for (Y in this case), press
[SOLVE] (which is [ALPHA][ENTER] on the TI-83). An approximate value of
Y which solves the equation will appear on the Y = line. To find the other
values of Y, simply change the guess and bounds to appropriate values and
press [SOLVE].
Using your solver, verify that the three values of Y are indeed Y =
-3, 1 and 2 (which you can also verify by hand).
Theoretically, we could take the equation y^3
- x*y = -6 and solve it for y. This would yield 3 formulas expressing
y in terms of x (because cubics have 3 solutions), i.e. we can view y as
a function of x (or many functions of x), as in y = x^2 or something. In
this case, we could actually solve the equation for y in terms of x (with
the help of high powered Mathematics), but the answers are extremely complicated,
too complicated to be of much value to us at this point.
From a graphical perspective however, the 3 functions
are easy to see. If we limit our x-values to those very close to x = 7,
the plot of all pairs (x,y) with y^3 - x*y = -6, looks like 3 separate
functions (which I've named f, g and h):
![[Maple Plot]](images/impdiff6.gif)
Near x = 7, which of f '(x), g'(x) and h'(x) are
positive or negative? Which is closest to zero?
Using the zoomed in view below, estimate f '(7)
and g'(7). You should probably draw in a tangent line.
![[Maple Plot]](images/impdiff7.gif)
Picking various values of x near 7, and solving
for y values near -3 (so we're "on the graph of h(x)"), gives:
|
x
|
y
|
|
6.98
|
-2.996998950
|
|
6.99
|
-2.998499737
|
|
7.0
|
-3.0
|
|
7.01
|
-3.001499738
|
|
7.02
|
-3.002998950
|
Use the table above to estimate dy/dx at the point
(7, -3).
You should now have decent estimates of dy / dx
at each of the points (7, -3), (7,1) and (7,2) (i.e., h'(7), g'(7) and
f '(7)). Let's see how we can get all 3 of these quickly using symbolic
differentiation and the chain rule! The key idea is to think of y not as
a variable, but as a function of x, say f(x). Our original equation can
then be viewed as [f(x)]^3 - x*f(x) = 6. The left side is some function
of x and so is the right side. These two functions (the left hand side
of the equation and the right hand side of the equation) must be equal,
and hence their derivatives are also equal. This says that
I can handle the right side derivative, it is 0. Use the chain rule and
product rule to calculate the derivative of the left hand side (your answer
will involve both f(x) and f '(x)).
The resulting equation "your left hand side derivative" = "my right
hand side derivative" (which is zero in this case), can normally be solved
for f '(x) with the following simple strategy:
-
Move all terms with an f '(x) in them to the left, and all terms without
f '(x) to the right side of the equal sign;
-
Factor out f '(x) on the left and divide through by the other factor.
If you now replace f '(x) with dy/dx, and f(x) with y, you have a formula
(in terms of both x and y) for the slope of the curve defined by the original
equation at the point (x,y). Do this, then substitute in the three pairs
x = 7, y = -3, then x = 7, y = 1 and finally x = 7, y = 2, to see if your
symbolic values for dy/dx agree with your estimates for slopes above.
dy/dx at (7,2) =
our guess was:
dy/dx at (7,1) =
our guess was:
dy/dx at (7,-3) =
our guess was:
Once you become comfortable with thinking of y as f(x), you needn't
actually replace y by f(x). Just realize that each time you differentiate
something with a y in it, you need to use the chain rule and "multiply"
by y' or dy/dx, or the product rule etc. Here is a sample problem done
both ways. Consider sin(x) + y*e^x = y^2. Here is a plot of a piece of
the graph of all pairs (x,y) which satisfy this equation.
The plot suggests that both (0,0) and (0,1) satisfy this equation. Indeed,
sin(0) + 0*e^0 = 0^2 and sin(0) + 1*e^0 = 1^2, so these points are on the
graph. The slope at (0,0) is negative. The slope at (0,1) is positive.
Thus, we would expect dy/dx to be negative at (0,0) and positive at (0,1).
Replacing y by f(x) in our equation yields:
sin(x) + f(x)*e^x = [f(x)]^2.
To differentiate the left hand side, we need the product rule for f(x)*e^x.
The left hand side has derivative
cos(x) + f '(x)*e^x + f(x)*e^x.
The chain rule is needed on the left, and the derivative of [f(x)]^2 is
2*f(x)*f '(x). Equating the two derivatives gives
cos(x) + f '(x)*e^x + f(x)*e^x = 2*f(x)*f '(x).
Moving terms with f '(x) in them left and others right yields
f '(x)*e^x - 2*f(x)*f '(x) = -cos(x) - f(x)*e^x.
Factoring out f '(x) on the left gives
f '(x) * [e^x - 2f(x)] = -cos(x) - f(x)*e^x.
Dividing through by the other factor of the left side yields
f '(x) = (-cos(x) - f(x)*e^x) / (e^x - 2f(x) ) or (dy/dx) = (-cos(x)
- y*e^x) / (e^x - 2y).
Putting in x = 0 and y = 0 yields dy/dx = -1 / 1 = -1, a negative value.
For the point (0,1), we get dy/dx = -2 / -2 = 1, a positive value. Things
look good.
If we simply view y as a function of x, we can save some time. However,
we need to realize that y*e^x requires the product rule, and y^2 requires
the chain rule (since y is really some f(x)). Leaving y as is, the equation
sin(x) + y*e^x = y^2, upon differentiating both sides, gives
cos(x) + (dy/dx)*e^x + y*e^x = 2*y*(dy / dx).
Moving terms with dy/dx left and those without dy/dx right produces
(dy/dx)*e^x-2y(dy/dx) = -cos(x) - ye^x.
Factoring dy/dx out on the left and division through by the other factor
gives
dy/dx = (-cos(x) - ye^x) / (e^x - 2y),
the same result as before. It's perfectly fine to always replace y by f(x)
and use the first method, so long as you always convert back to y and y'
(or y and dy/dx), instead of f(x) and f '(x) respectively.