The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Math Topics »

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: derivatives form...
Replies: 1   Last Post: Dec 27, 2012 8:07 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 132
Registered: 11/27/12
Re: derivatives form...
Posted: Dec 27, 2012 8:07 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

What you have written is non-sense. f'(x- 2) MEANS the derivative of f evaluated at x- 2 but you have given no formula for f.

What I THINK you mean is "if f(x)= x- 2 what is f'(x)?". That's a much easier question. In fact, it follows immediately from the fact that the derivative is a generalization of the SLOPE of a line. Since f(x)= x- 2 is a linear function, its derivative IS its slope. What is its slope?

Another way to do this would be to use the limit definition of the derivative: f'(x)= lim(as h goes to 0) of (f(x+h)- f(x))/h. With f(x)= x- 2, f(x+h)= (x+h)- 2.
f(x+h)- f(x)= (x+ h- 2)- (x- 2)= h so (f(x+ h)- f(x))/h= 1 so the limit is 1.

Yet another way is to use the "theorems" derived from that formula. In particular, (f+ g)'= f'+ g'. (cf)'= cf' (where c is a constant) and (x^n)'= nx^(n-1). Here f= x and g= -2. f= x^1 so f'= 1(x^0)= 1. g= -2= -2x^0 so g'= -2(0)x^{-1}= 0.

If f(x)= 2x^2, the first idea doesn't help because it is not linear. But we can still say f(x+h)= 2(x+h)^2= 2(x^2+ 2hx+ h^2)= 2x^2+ 4xh+ 2h^2. Then f(x+h)- f(x)= 2x^2+ 4xh+ 2h^2- 2x^2= 4xh+ 2h^2 so that (f(x+h)- f(x))/h= (4xh+ 2h^2)/h= 4x+ 2h. What is the limit of that as h goes to 0?

Or use the theorem that the derivative of x^2 is 2x^(2-1)= 2x.

Date Subject Author
Read derivatives form...
Read Re: derivatives form...

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.