Implicit DifferentiationDate: 10/21/96 at 22:15:0 From: Alan Zielinski Subject: Implicit Differentiation How do I find dy/dx for: ((xy)^(1/2)) + ((x + 2y)^(1/2)) = 4 Date: 10/22/96 at 2:9:55 From: Doctor Pete Subject: Re: Implicit Differentiation You wish to find dy/dx, that is, the rate of change in y as a function of x. Some functions aren't always expressible in the form y = f(x), as the above example shows. Rather, the best we can do is to write it as some f(x,y) = 0. But this doesn't mean that we can't find derivatives -- in the latter case, we see that the existence of f(x,y) means that x and y are related in some sense; changing x forces y to change as well. This is the idea behind implicit differentiation. If you understand the concept well enough, you should see that it is only a generalization of "regular" differentiation. Here is an example: Find dy/dx if f(x,y) = xy + x^2 - y^2 + Sqrt[x-y] = 0. (Here, Sqrt[ ] is the square root function.) First find df(x,y)/dx; that is, take d/dx of the above (the right-hand side remains zero): df(x,y) d ------- = ---- (xy + x^2 - y^2 + Sqrt[x-y]) dx dx dy dy 1 1 dy = x ---- + y + 2x - 2y ---- + --- --------- (1 - ----) dx dx 2 Sqrt[x-y] dx 1 dy 1 = (x - 2y - -----------) ---- + 2x + y + ----------- = 0. 2 Sqrt[x-y] dx 2 Sqrt[x-y] This was done by taking the derivative as usual, except when there was a y term, I wrote dy/dx. (Notice that d[y^2]/dx = 2y * dy/dx; this is actually from the chain rule.) The last step was simply collecting terms. So to find dy/dx, the rest is easy; simply move the terms without dy/dx to the other side, and divide by x - 2y - 1/(2 Sqrt[x-y]). The result is actually a new function g(x,y). So I'll carry out the first few steps on your example, just to start you off: d ---- (Sqrt[xy] + Sqrt[x+2y] - 4) = 0 dx 1 1 dy 1 1 dy --- -------- ( x ---- + y ) + --- ---------- ( 1 + 2 ---- ) - 0 = 0 2 Sqrt[xy] dx 2 Sqrt[x+2y] dx Now collect the dy/dx terms, move the other terms to the other side and solve for dy/dx. Hope this helps, -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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