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Homework Help: Differentiate -5/3x

  1. Jun 14, 2008 #1
    i can differentiate most other simple functions.. .though can someone please help me to understand why the derivative of f(x)=-5/3x is simply 5/3x^2?
     
  2. jcsd
  3. Jun 14, 2008 #2

    Doc Al

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    Do you know the power rule for differentiating functions like [itex]x^n[/itex]? (Hint: [itex]1/x = x^{?}[/itex].)
     
  4. Jun 14, 2008 #3

    Kurdt

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    Its basically following the rule:

    [tex] \frac{d}{dx} ax^n = anx^{n-1} [/tex]

    where in your example [itex]a=\frac{-5}{3}[/itex] and [itex] n = -1[/itex].

    EDIT: puppy interupted hence late reply :tongue:
     
    Last edited: Jun 14, 2008
  5. Jun 14, 2008 #4
    1/x = x^-1 and yes i can differentiate functions in the form x^n
     
  6. Jun 14, 2008 #5

    Doc Al

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    Then you should be able to solve this as that's all that's needed.
     
  7. Jun 14, 2008 #6
    oh i see , i just didnt realize n was -1, but obviously it is wen its on the denominator, problem solved. cheers
     
  8. Jun 14, 2008 #7
    F(x) = [tex]\frac{-5}{3x}[/tex]

    The first thing you got to do is re-write the function so we get:

    F(x) = [tex]\frac{-5}{3}[/tex] . [tex]\frac{1}{x}[/tex]

    F(x) = [tex]\frac{-5}{3}[/tex] . x[tex]^{-1}[/tex]

    after re-writing the function, find the most appropriated way to derivate the function in this case it would be: [tex]\frac{d}{dx}[/tex] [u[tex]^{n}[/tex]] = nu[tex]^{n-1}[/tex] u'

    Using this rule we get:

    F(x) = [tex]\frac{-5}{3}[/tex] . x[tex]^{-1}[/tex]

    F'(x) = (-1) . [tex]\frac{-5}{3}[/tex] . x[tex]^{-2}[/tex] (1) , then simplify

    F'(x) = [tex]\frac{5}{3}[/tex] . x[tex]^{-2}[/tex], then re-write

    F'(x) = [tex]\frac{5}{3}[/tex] . [tex]\frac{1}{x^{2}}[/tex]

    F'(x) = [tex]\frac{-5}{3x^{2}}[/tex]
     
  9. Jun 14, 2008 #8

    HallsofIvy

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    You don't have to use the power rule for negative powers, you can use the quotient rule: -5/3x= f(x)/g(x) with f(x)= -5 and g(x)= 3x.
    (f/g)'= (f'g- fg')(g2)

    Since f'= 0 and g'= 3, that gives (-5/3x)'= ((0)(3x)- (-5)(3))/(9x2= 5/3x2.

    The reason I mention that is that before you can use the power rule for negative powers you have to prove it for negative powers- and you do that by using the quotient rule:
    x-n= 1/xn= f/g with f(x)= 1, g(x)= xn. f'= 0, g'= nxn-1 so (x-n)'= (1/xn)'= ((0)(xn)- 1(nxn-1)/x2n= n xn-1/x2n= n x(n-1)- 2n= n x-n-1.
     
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