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Differentiate -5/3x

  • Thread starter stat643
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  • #1
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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?
 

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  • #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].)
 
  • #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:
  • #4
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1/x = x^-1 and yes i can differentiate functions in the form x^n
 
  • #5
Doc Al
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1/x = x^-1 and yes i can differentiate functions in the form x^n
Then you should be able to solve this as that's all that's needed.
 
  • #6
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oh i see , i just didnt realize n was -1, but obviously it is wen its on the denominator, problem solved. cheers
 
  • #7
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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]
 
  • #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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