Derive x/(3x-1) using first principles

  • Thread starter Zipzap
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  • #1
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Ok, I am really stuck on this one. I tried using the first principles formula and everything, but I don't get two separate terms like I am supposed to in the actual derivative. I always end up with -1/(3x-1)^2 when I try. Can someone please help me out?
 

Answers and Replies

  • #2
35,287
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Ok, I am really stuck on this one. I tried using the first principles formula and everything, but I don't get two separate terms like I am supposed to in the actual derivative. I always end up with -1/(3x-1)^2 when I try. Can someone please help me out?

Show us what you've done. It should start off something like this:
[tex]\lim_{h \to 0}\frac{\frac{x + h}{3(x + h) - 1} - \frac{x}{3x - 1}}{h}[/tex]
 
  • #3
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I guess it is odd to end up with the right answer.
 
  • #4
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Well, Mark44, I have that in the beginning to. I then multiply denominators to get a common one, and I end up with something like:

{ [ (x+h)(3x-1) - x(3x + 3h -1) ] / [ (3x+3h-1)(3x-1) ] } / h

{ ( [3x^2 - x + 3hx - h] - [3x^2 + 3hx -x] ) / [ (3x+3h-1)(3x-1) ] } / h

My problem is that everything cancels out except for h, and that leaves me with only one term, which I know is incorrect. What am I doing wrong here?
 
  • #5
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If you write this as x*(3x-1)-1 and do the chain rule you get (3x-1)-1-x*(3x-1)-2*3 which if you put (3x-1)-2 in both denominators gives you (3x-1)/(3x-1)2-(3x)/(3x-1)2 and the 3x cancels out which gives what you got so you got the right answer.
 

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