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Calculus made easy by Thompson: exercise vii: 1

  1. Sep 4, 2013 #1
    Hello, I require some assistance with this problem. It reads:

    If, u=1/2x^3; v=3(u+u^2); w=1/v^2;

    Find dw/dx.

    It's clear that you should use the chain rule so I begin by calculating the derivative for each one.

    I get v^(-2); 6u+3; and 3/2x^2.

    I then have to multiply them but this is where the problem comes. The answer I derived was (3x^(2)-12u-6)/2v^3

    This doesn't seem right and the answer say it isn't at the back of book.

    Is the problem the multiplication?
     
  2. jcsd
  3. Sep 4, 2013 #2

    Mentallic

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    v-2 = what?
    6u+3 = what?
    etc.
     
  4. Sep 4, 2013 #3
    In order of appearance: dw/dv then dv/du then du/dx.
     
  5. Sep 4, 2013 #4

    HallsofIvy

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    The first is incorrect. [itex]w= 1/v^2= v^{-2}[/itex] itself. The derivative is [itex]w'= -2v^{-3}[/itex]
    but apparently that is a typo since you have [itex]v^3[/itex] below.
    (u= (1/2)x^3, not 1/(2x^3)?)

    Specifically, you have [itex](-2v^{-3})(6u+ 3)(3/2)x^2[/itex]. With [/itex]u= (1/2)x^3[/itex], [itex]6u+ 3= 3x^3+ 3[/itex] so this is -9(x^5+ 3x^2)/v^3

    What is the answer in the book? Do they have x, u, and v or have the converted everything to x?
     
  6. Sep 4, 2013 #5
    Everything is x.
     
  7. Sep 4, 2013 #6
    It reads: -(3x^2(3+3x^3))/(27((1/2)x^3 + (1/4)x^6)^3)
     
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