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Chain Rule (with fractions)

  1. Aug 26, 2009 #1
    Okay, I know how to differentiate regular functions. But when it comes to fractions, I'm hopeless. This may be an extremely simple one to some, here is the function; "1/4x-7"
    I have to differentiate that using the chain rule.

    I think that u=4x-7, but im not sure. As i said, im horrible when it comes to fractions in the chain rule.
     
  2. jcsd
  3. Aug 26, 2009 #2

    HallsofIvy

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    First, what exactly is your function? Is it (1/4)x- 7 or 1/(4x)- 7 or 1/(4x- 7)?

    Since you say "I think that u=4x-7", I assume it is 1/(4x- 7). As for what "u" is, that's your choice. TRY something and see if it works. If you choose u= 4x- 7 then 1/(4x-7)= 1/u= u-1. Can you differentiate that? And you certainly ought to be able to differentiate u= 4x-7 (the derivative of a linear function is just its slope). Finally, the chain rule says
    [tex]\frac{df}{dx}= \frac{df}{du}\frac{du}{dx}[/tex].
     
  4. Aug 26, 2009 #3
    well the function is actually 1/4x-7. there is no brackets so i naturally assumed u=4x-7. I get it how the answer is -4/(4x-7)2 but why is it squared? i was thinking 1/u X -4 where u=4x-7. What I want to know now is why is the u squared?
     
  5. Aug 26, 2009 #4
    Ohh! when there is a power of at the bottom of the fraction, does it go up? like from to the power of one it'll go up to the power of 2?
     
  6. Aug 26, 2009 #5

    Mark44

    Staff: Mentor

    I'm guessing that your function is written in your text or worksheet like this:
    [tex]\frac{1}{4x - 7}[/tex]

    When you write it on a single line, you have to put parentheses around the terms in the denominator. IOW, like this: 1/(4x - 7).

    The way you wrote it, without parentheses, would be interpreted like this:
    [tex]\frac{1}{4} x - 7[/tex]

    BTW, why did you post this under Precalculus Mathematics? This is obviously a calculus problem.
     
  7. Aug 26, 2009 #6

    HallsofIvy

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    [tex]\frac{1}{x^2}= x^{-2}[/tex] and the derivative of [itex]x^n[/itex] is [itex]nx^{n-1}[/itex] where n is any number [As GibZ pointed out- any number except -1!]. If n= -2 what does that give you?

    (You could also do that problem using the "quotient rule": the derivative of u(x)/v(x)= (u'v- uv')/v2. If u=1 and v= x2, what does that give you?
     
    Last edited: Aug 27, 2009
  8. Aug 26, 2009 #7
    Well my teacher told me that this is sort of like the introduction to calculus.
     
  9. Aug 27, 2009 #8

    Gib Z

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    Careful! Not if n=0 !. =]

    TheAkuma: Introduction or not, please post under the calculus section. If helps the homework helpers out if they know what kind of things to expect to help people with.
     
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