Chain Rule: Function Notation

Date: 9/5/96 at 22:7:25
From: Jason Ramsey
Subject: The Chain Rule

How can I apply the chain rule to find du/dx for 
dy/dx=(nu^n-1)*(du/dx)?

Thank you
Jason Ramsey

Date: 9/6/96 at 16:20:8
From: Doctor Mike
Subject: Re: The Chain Rule

Hello Jason, 
  
I think I can help you out, but first, let me make sure we are
both talking about the same thing.  A well-known derivative fact
is that if y=x^n then y' or dy/dx is n*x^(n-1).  I suppose you
are talking about when y is not just "x" to the n, but is the
more complicated "u(x)" to the n, where u(x) is another function.
If you are really asking about something else, just write again.
  
When "u" is a function of x like u(x)=sin(x) or u(x)=x^4+5x+2 , 
then the particular instance of the chain rule you are writing
about will tell you how to find derivatives of y=(u(x))^n .  The
du/dx part could also be written u'(x) and it is the derivative
of u(x) with respect to x.  For the 2 examples of mine, we get : 

       dy/dx  =  n*(sin(x))^(n-1)*cos(x) 
  
       dy/dx  =  n*(x^4+5x+2)^(n-1)*(4x^3+5)
  
since the derivative of sin is cos, and the derivative of my 4th
degree polynomial is 4x^3+5 .
  
Personally, I prefer the function notation for the chain rule,
namely that if c(x) is a composition of functions f(x) and u(x)
defined by c(x)=f(u(x)), then the derivative c'(x) of c is just
f'(u(x))*u'(x).  That is, the derivative of the whole function
is the derivative of the "outside" function, evaluated at the 
"inside" function, times the derivative of the "inside" function.
For your example the "outside" function is f(x)=x^n, and its
derivative is f'(x)=n*x^(n-1), and the "inside" function could 
be either of my u(x) examples or any other differentiable function. 

I hope this helps.  

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