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/
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