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Checking My Method For Differentiation

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Differentiate the following with respect to x

    y = [itex]\frac{4}{x^{3}}[/itex] + [itex]\frac{x^{3}}{4}[/itex]


    3. The attempt at a solution

    So the problem here is really getting this into a form that is easy to differentiate and i'd just like to show what i'm doing before I go ahead and do the rest of my questions. The step i'm least confident with is applying chain rule to each term separately...

    I want to use the chain rule here, so I would rearrange to this;

    y = (x3)-4 + (4)-x3

    dy/dx = -4(x3)-53x2 - x3(4)-x3-1.0

    dy/dx = -4(x3)-53x2

    dy/dx = -12x2(x3)-5

    (made irrelevant by discussion below)

    So, is this ok? I am concerned that my second term disappears, but that's what it looks like I have to do.

    Thanks for any help you can give!
     
    Last edited: Oct 22, 2013
  2. jcsd
  3. Oct 22, 2013 #2

    FeDeX_LaTeX

    User Avatar
    Gold Member

    How did you get that?
     
  4. Oct 22, 2013 #3
    [itex](x^{3})^{-4}[/itex] = [itex]4/x^{3}[/itex] is it not?
     
  5. Oct 22, 2013 #4

    FeDeX_LaTeX

    User Avatar
    Gold Member

    It is not -- ##(x^3)^{-4} = x^{-12}##.

    ##\frac{4}{x^3} = 4x^{-3}##
     
  6. Oct 22, 2013 #5

    Mark44

    Staff: Mentor

    No, it is not.

    ##\frac 4 {x^3} = 4x^{-3}##

    For your problem there is no need for the chain rule.
     
  7. Oct 22, 2013 #6
    I have made this mistake a few times before... Thanks for catching it.
     
  8. Oct 22, 2013 #7
    That's become somewhat clearer to me after my mistake was pointed out.

    so y = [itex]4x^{-3} + x^{3}(4^{-1})[/itex]

    dy/dx = [itex]-12x^{-4} -4^{-2}3x^{2}[/itex]

    dy/dx = [itex]-12x^{-4} - (3/2) x^{2}[/itex]
     
  9. Oct 22, 2013 #8

    Mark44

    Staff: Mentor

    y = 4x-3 + (1/4)x3
    dy/dx = -12x-4 + (3/4)x2

    You are using the power rule on something that is not a function. 4-1 is a constant, so its derivative is zero.
     
  10. Oct 22, 2013 #9
    Ah, I understand.

    Thank you.
     
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