Chain Rule: Function NotationDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/