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Two Ways to Differentiate

Date: 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

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

    x * -- (x) = x * 1 = x ,

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

    x^2 * -- (1) = x^2 * 0 = 0 .

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/   
Associated Topics:
High School Calculus

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