Chain RuleDate: 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/ |
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