Chain Rule

Date: 10/23/96 at 16:17:51
From: Zach Robinson
Subject: Math Problem

I have a question on the "chain rule" when finding the derivatives of
polynomials.  

One such problem is:    f(x) = (x-1)(x-1)-2x
Where (x-1) is squared.
How would you find the derivative of this problem?

Thank you,

Zachary Robinson

Date: 10/23/96 at 21:0:47
From: Doctor Charles
Subject: Re: Math Problem



The chain rule of differentiation says that if you have two functions 
f and g and you let z=g(y) and y=f(x) (in other words z=g(f(x))) then
the derivative of z with respect to x, (written dz/dx) is:

        dz/dx = dz/dy * dy/dx.

So it *looks* like you are cancelling the dy 's in the multiplication 
of the two fractions. It is important to realise that this is only 
what it looks like and is not a reason for doing it as the terms dx, 
dy and dz have no meaning when they are on their own and not in a 
fraction.

This is why a lot of people prefer the function notation of writing 
a ' after the letter denoting the function to signify its derivative.

  e.g     if h(x) = x^3  then  h'(x) = 3 x^2.

So in the other notation the chain rule says:

   If h(x) is the function g(f(x)) then

      h'(x) = g'(f(x)) * f'(x)

So putting this into practice with your example...

       f(x) = (x-1)^2 - 2x

First of all note that we can write f(x) as f(x) = h(x) + k(x)
where I am letting h(x) = (x-1)^2 and k(x) = -2x.

Then (not the chain rule but a simpler result) says that:
   f'(x) = h'(x) + k'(x)

so first we find h'(x). This is where the chain rule comes in.

h(x) = g(f(x)) where f(x) = x - 1 and g(y) = y^2

f' and g' are easy to work out:
f'(x) = 1 and g'(y) = 2y   (in our case we put y = x - 1)

we then write down what g'(f(x)) * f'(x) is:

h'(x) = g'(f(x)) * f'(x) = 2(x-1) * 1 

so h'(x) = 2(x-1) and it is easy to work out that k'(x) = -2

so (finally!)

f'(x) = h'(x) + k'(x) = 2(x-1) - 2 = 2x - 4

I hope that you can follow this. I don't know which notation you are 
used to but from the formation of the question this seemed likely to 
be the most helpful.

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