Results 1 to 5 of 5

Math Help - Supposedly easy derivative:

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    11

    Supposedly easy derivative:

    Hello all,

    I'm in charge of finding the derivative of 1 \over 4x-3.

    I first used the quotient rule, which came out to be

     0 * (4x-3) - ((4x-3) * 1) \over (4x-3)^2

    and simplified to

    - 4x-3 \over (4x-3)^2


    However, this isn't complete. I understand that I need to use the chain rule on the numerator next, according to Wolfram-Alpha, but I don't understand why I need the chain rule since there's only 1 function. Can anyone shed some light on this?

    Thank you in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Use the law of exponents that says \dfrac{1}{a} = a^{-1} to rewrite \dfrac{1}{4x-3} = (4x-3)^{-1}

    You can the use the chain rule
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2011
    Posts
    11
    Oh wow! Haha.

    Now that you show that, it's pretty easy to see. I'd imagine I should do this for all fractions with 1 in the numerator correct?

    It makes complete sense:

     (4x-3)^{-1}

    (4x-3)^{-1} = (4x-3)^{-2} * 4

     4 \over (4x-3)^2


    EDIT: I should say, a constant in the numerator
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Yes, but bear in mind that it will not be affected by the exponent. That is \dfrac{2}{x-1} = 2(x-1)^{-1} \neq (2(x-1))^{-1}


    Edit. Your answer is missing a minus sign out front.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,454
    Thanks
    1868
    Quote Originally Posted by sinx View Post
    Hello all,

    I'm in charge of finding the derivative of 1 \over 4x-3.

    I first used the quotient rule, which came out to be

     0 * (4x-3) - ((4x-3) * 1) \over (4x-3)^2
    No, that's not correct. The quotient rule says that
    \left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}
    Here f(x)= 1 so f'(x)= 0, g(x)= 4x- 3 so that g'= 4.
    \frac{0(4x-3)- (4)(1)}{(4x- 3)^2}

    and simplified to

    - 4x-3 \over (4x-3)^2


    However, this isn't complete. I understand that I need to use the chain rule on the numerator next, according to Wolfram-Alpha, but I don't understand why I need the chain rule since there's only 1 function. Can anyone shed some light on this?

    Thank you in advance!
    No, not "use the chain rule on the denominator". But you did forget to differentiate 4x- 3.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig Sub to Integration by Parts (Supposedly)
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 8th 2010, 09:59 PM
  2. easy derivative question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 18th 2009, 08:07 PM
  3. Replies: 0
    Last Post: October 26th 2009, 08:34 PM
  4. VERY EASY derivative question
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 30th 2009, 02:01 PM
  5. Replies: 3
    Last Post: June 10th 2009, 01:26 PM

Search Tags


/mathhelpforum @mathhelpforum