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Implicit Differentiation


Date: 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/   
    
Associated Topics:
High School Calculus

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