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Implicit Differentiation


Date: 10/15/2001 at 13:05:50
From: Gary Richardson
Subject: Implicit differentiation

We were given the equation 1 - xy = x - y. We found y' by implicit 
differentiation and got: y' = (1+y)/(1-x). However, if we solve the 
equation for y we get: y = -1 and so y' = 0.

We have not been able to show that (1+y)/(1-x) = 0.

We feel certain that solving for y first is a valid approach, but 
we're confused because we did not get the same value for y' both ways!

Where have we gone wrong?


Date: 10/15/2001 at 14:17:58
From: Doctor Peterson
Subject: Re: Implicit differentiation

Hi, Gary.

Actually, you haven't gone wrong; you just missed the last step.

By implicit differentiation, you found that y' = (1+y)/(1-x).

Then by solving the equation for y, you found that y = -1. That is, 
your equation is really just a horizontal line, and y' = 0.

Now take your equation for y' and plug in y = -1. You get y' = 0, just 
as you wanted!

Actually, there's another dimension to this problem that you missed. 
When you solved for y, you really got

        1-x
    y = ---
        x-1

which you then simplified by canceling x-1. But you can't do that if 
x = 1. A better solution would be

            1 - xy = x - y

    xy + x - y - 1 = 0

    (x - 1)(y + 1) = 0

                 x = 1 OR 
                 y = -1

So really your equation describes a pair of crossed lines, a 
horizontal line at y = -1 and a vertical line at x = 1. This has slope 
0 on the former, and undefined slope on the latter. Note that y is 
nothing like a function of x, so solving for y doesn't accomplish 
everything you hoped for.

Now try plugging in x = 1 in your implicit derivative. Everything will 
work out just right.

It's important to realize that when you take the derivative 
implicitly, it is assumed that x and y satisfy the original equation; 
when you plug in any pair that does, you will get the derivative at 
that point. Here, knowing only x = 1 or only y = -1 determines the 
point, and therefore the derivative. So your formula for the 
derivative did not have to equal 0 everywhere; it just had to be zero 
when y = -1, and undefined when x = 1, which is true.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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