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Chain Rule and Product Rule: Differentiating

Date: 02/19/98 at 18:18:27
From: Kelly
Subject: Differentiating w/ chain rule and product rule

To differentiate:
   x^2 * (2-3x)^7

Date: 02/19/98 at 21:56:59
From: Doctor Sam
Subject: Re: Differentiating w/ chain rule and product rule


Well, there are a couple of things going on here. One is the Product 
Rule and the other is the Chain Rule. They are often used together 
(as in your example) but they don't have to be. That may be part of 
your confusion.

Let's first look at the Product Rule and not worry about the Chain 

The Product Rule tells how to find the derivative of the product of 
two functions.  

         (f(x) g(x))' = f'(x)g(x) + f(x)g'(x).  

If, for example, f(x) = x^2 -2x and g(x) = 3x^2 + 4x and you need the 
derivative of (x^2 -2x)(3x^2 + 4x), you can get it by finding:

         (x^2 -2x)'(3x^2 + 4x) + (x^2 -2x)(3x^2 + 4x)' =
         (2x - 2) (3x^2 + 4x) + (x^2 -2x)(6x + 4)


Now lets look at the Chain Rule by itself.  The Chain Rule shows how 
to find the derivative of functions that have been combined by 
      for example, (2x+7)^2
      which is a composition of f(x) = 2x+7 and g(x) = x^2.

To find the derivative of (2x+7)^2 = g(f(x)), the Chain Rule says 

      "differentiate the outside funtion"  
         [that's g and g'(x) = 2x] 

      then "differentiate the inside function"  
         [that's f and f'(x) = 2]  

      and finally, "multiply g'(f(x)) times f'(x)"  
         [that's 2(2x+7) times 2].

In symbols these words are usually written as 

         g(f(x)]' = g'(f(x))f'(x).

And that looks really confusing. I'm not surprised that you don't have 
any good intuitions about this yet.

Of course you did it with a substitution. If we write u = 2x+7 then 
we are trying to differentiate (2x+7)^2 which is u^2. This derivative 
is just 2u but you have to substitute back to get 2(2x+7). You then 
multiply by u' = 2 to get the same answer.

In symbols this is usually written as y = u^2 and u = 2x+7. Instead of 
writing y' we write dy/du, and instead of u' we write du/dx.

The original problem was to find the derivative of y = (2x+7)^2 where 
y is a function of x. (In order to compute a value of y you need a 
value of x.)  This derivative is called dy/dx.  

You introduced the helping variable u and got y as a function of u 
and u as a function of x. This forms a CHAIN of functions from 
x to u to y. In order to compute a value of y from a value of x you 
must first figure out what u is and the use it to find y. In order to 
find the derivative dy/dx of the complicated function (2x+7)^2 we use 
the easier functions that are chained together:

        x ------->   u  --------> y
to find dy/dx (which means roughly "how fast y changes as x changes") 
you must first figure out how fast u changes when you change x and 
then figure out how fast y changes when you change u. The Chain Rule 
says, in symbols, that 

        dy/dx = (dy/du)(du/dx) 
You multiply the two derivatives to get the derivative that you want.

Now to your problem that uses both the Product Rule and the Chain 

To find [ x^2 * (2-3x)^7 ]'  first use the product rule:

        [ x^2 * (2-3x)^7 ]' = [x^2]' * [(2-3x)^7]  + [x^2][(2-3x)^7]' 

The first derivative we can do directly:  [x^2]' = 2x

The second one needs the Chain Rule.  To find [(2-3x)^7]', let
             u = 2-3x and y = u^7.

The Chain Rule says that 

                      dy/dx = (dy/du) (du/dx)  

                      dy/du = 7u^6 = 7(2-3x)^6

                      du/dx = -3

                   so dy/dx = 7(2-3x)^6 (-3) or -21(2-3x)^6.

Now put it all together:   

        [ x^2 * (2-3x)^7 ]' = 2x[(2-3x)^7]  + x^2[-21(2-3x)^6]

I hope that helps!

-Doctor Sam,  The Math Forum
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Associated Topics:
High School Calculus

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