1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Find the Derivative/Simplification of 2 Rational Expressions

  1. Jul 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of:
    y = [(2x - 5)^4][(8x^2 - 5)^-3]

    2. Relevant equations
    I get:
    y' = -[(48x(2x - 5)^4) / (8x^2 - 5)^4] + [(8(2x - 5)^3) / (8x^2 - 5)^3]

    Wolfram gets:

    3. The attempt at a solution
    I do everything Wolfram does here:
    I just don't know how it simplifies what you see at the bottom (of the "Show Steps" section that is) to get the solution you see at the top.
  2. jcsd
  3. Jul 14, 2012 #2


    User Avatar
    Homework Helper

    You should have learned how to add/subtract rational expressions in algebra. Just like in numerical fractions, you can only add/subtract if the denominators are the same. If they are not, you need to find the LCD (least common denominator) first. What is the LCD in this case?
  4. Jul 14, 2012 #3


    User Avatar
    Science Advisor

    [tex]y'= -\frac{48x(2x- 5)^4}{(8x^2- 5)^4}+ \frac{8(2x- 5)^3}{8x^2- 5)^3}[/tex]

    You have two fractions, one with denominator [itex](8x^2- 5)^4[/itex], the other with denominator [itex](8x^2- 5)^3[/itex]. You can first simplify the calculation by factoring [itex](2x- 5)^3[/itex] out of the numerators and [itex](8x^2- 5)^3[/itex] out of the denominators:
    [tex]\frac{(2x-5)^3}{(8x^2-5)^3}\left(\frac{-48x(2x-5)}{8x^2-5}+ 4\right)[/tex]
    Now, get a common denominator by multiplying that "4" by [itex](8x^2-5)/(8x^2- 5)[/itex].
    [tex]\frac{(2x-5)^3}{(8x^2- 5)^3}\left(\frac{-48(x(2x-5)}{8x^2- 5}+ \frac{4(8x^2- 5}{8x^2- 5}\right)[/tex]
    [tex]=\frac{(2x-5)^3}{(8x^2- 5)^3}\left(\frac{-96x^2+ 240+ 32x^2- 20}{8x^2- 5}\right)[/tex]
    [tex]= \frac{(2x-5)^3}{(8x^2- 5)^3}\left(\frac{220- 64x^2}{8x^2- 5}\right)[/tex]
    Last edited by a moderator: Jul 15, 2012
  5. Jul 14, 2012 #4
    Okay then. I understand. Thank you both.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook