Results 1 to 2 of 2

Math Help - Quiotent rule

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    4

    Quiotent rule

    I understand all the steps in solving the problem except for the final answer.



    Y= (-2t^2+1) / (3t^2-5)



    I know that f (t) * g' (t) = (-12t^3+6t)

    g (t) * f' (t) = (-12t^3+20t)



    y= (14/3t^2-5)^2 (answer)

    I know that the denominator is always gonna be the second part of the equation squared. But how in the heck do you get the 14?
    Thanks for all the help Peace
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Let f(t) = -2t^{2} + 1 and g(t) = 3t^{2} - 5.

    y' = \frac{f'(t)g(t) - f(t)g'(t)}{\left[g(t)\right]^{2}}

    You found f'(t)g(t) = -12t^{3}+20t and f(t)g'(t) = -12t^{3} + 6t.

    So:

    y' = \frac{-12t^{3} + 20t - \left(-12t^{3} + 6t\right)}{\left(3t^{2} - 5\right)^{2}}

    Distribute the negative sign and you should see that it works out to your answer. I'm guessing you forgot the parentheses when substituting everything in.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] quick question on product rule and equality rule for logs
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: October 19th 2011, 07:29 PM
  2. Replies: 5
    Last Post: October 19th 2009, 01:04 PM
  3. Replies: 0
    Last Post: October 20th 2008, 07:12 PM
  4. Replies: 2
    Last Post: December 13th 2007, 05:14 AM
  5. Replies: 3
    Last Post: August 31st 2006, 09:08 AM

Search Tags


/mathhelpforum @mathhelpforum