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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:
    http://www.wolframalpha.com/input/?i=derivative+(2x+-+5)^4(8x^2+-+5)^-3


    3. The attempt at a solution
    I do everything Wolfram does here:
    http://www.wolframalpha.com/input/?i=derivative+(2x+-+5)^4(8x^2+-+5)^-3
    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

    eumyang

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    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

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    So
    [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: Jul 15, 2012
  5. Jul 14, 2012 #4
    Okay then. I understand. Thank you both.
     
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