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- Thread starter stat643
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- #2

Doc Al

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- #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:

[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:

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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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Then you should be able to solve this as that's all that's needed.1/x = x^-1 and yes i can differentiate functions in the form x^n

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- #7

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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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(f/g)'= (f'g- fg')(g

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

The reason I mention that is that before you can use the power rule for negative powers you have to

x

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