Differentiating

Date: 01/14/98 at 11:53:53
From: Stephen Hack
Subject: Differentiation

Please help me differentiate

(y^2) * cos (1/y) = 2x + 2y

Date: 01/30/98 at 16:27:04
From: Doctor Joe
Subject: Re: differentiation

Dear Stephen,

I shall not answer your question directly. I hope that you can do it 
yourself. What I shall do is to give you some advice while performing 
an example of this type of implicit differentiation.

Recall that if f is a function of y and y is also a function of x, 
then to differentiate f with respect to x, we need to apply the 
so-called chain-rule, right?

            d          d         dy
            -- f(y) = (-- f(y))* --
            dx         dy        dx

which can also be written as:
   
     df   df   dy
     -- = -- * --
     dx   dy   dx  

For instance, differentiating y^3*sin(1/y^2) with respect to x gives:

          {d/dy (y^3*sin(1/y^2))} * dy/dx

        = {3y^2 * sin(1/y^2) + y^3 * cos(1/y^2) * (-2/y^3)} * dy/dx

In a more complicated example, the next thing you would do would be to 
group the terms that have a factor of dy/dx together. To show you 
another example:

Differentiate with respect to x,
           x^2 + y^3 = y

We start differentiating implicitly with respect to x:

      2x + (3y^2 * dy/dx) = dy/dx
=>     dy/dx * (3y^2 - 1) = -2x
=>                  dy/dx = 2x/(1-3y^2)

After grouping the dy/dx terms together, we were able to isolate them 
on one side of the equation.

Note that the final expression of dy/dx usually involves a quotient. 
In the last example that I gave, the denominator is (1-3y^2).  Don't 
think that everything is fine. Sometimes you have to check to see 
whether your derivative makes sense at certain special points - in 
this case it is the value when y = sqrt(1/3) .

Now you can solve the problem you posed using the same methods I 
illustrated in my examples. You will go through the following steps:

1) differentiate each term with respect to x (and use the chain-rule   
   when a term has y's in it),

2) rearrange and group the terms to solve for dy/dx.

Feel free to mail me if in doubt. Regards.

-Doctor Joe,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/