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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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