Two Ways to DifferentiateDate: 6/30/96 at 4:16:1 From: Park Seung Ryul Subject: Two Ways to Differentiate I want to know following question: If y=x^2, then y'=2x. But, if y = x+x+...+x (x*x) then y'= 1+1+...+1 (1*x) = x Why? Date: 6/30/96 at 10:56:56 From: Doctor Pete Subject: Re: math The problem with what you wrote is that x is the variable you are differentiating with respect to. When you write y = x+x+x+x+...+x , <--x times--> you can't say y' = 1+1+1+1+...+1 , <--x times--> because x is not constant (it's not even an integer). When you write "x times" and find y', you are actually calculating d x * -- (x) = x * 1 = x , dx because it's like "taking out" one x. Think of it this way: Why can't you say, y = x^2 = x+x+...+x (x times) = (1+1+...+1) (x times) \ + (1+1+...+1) (x times) \ (x times) + . . . / + (1+1+...+1) (x times) / so y' = 0+0+...+0 (x^2 times) = 0 ? This time, what I actually calculated was d x^2 * -- (1) = x^2 * 0 = 0 . dx The correct way to do this is y = x^2 = x+x+x+...+x (x times) y' = (1+1+1+...+1) + (x) <-- 1 time! <--x times--> = x + x = 2x. This is from the product rule: if f(x) = u(x)*v(x), then f'(x) = u'(x)*v(x)+u(x)*v'(x). So what I did was I differentiated the "x times." It's very confusing, because what actually happened was I let u(x)=x, v(x)=x. I hope this makes sense! -Doctor Pete, 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}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/