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Why the formula for derivatives

  1. Mar 16, 2005 #1
    The formula for finding a derivative for x^n is nx^(n-1) and the anti derivative is 1/(n+1) x^(n+1)
    Why is this the formula?
  2. jcsd
  3. Mar 16, 2005 #2
    If you take the derivative of a function using the (derivative) power rule, then you can always reverse it using the antiderivative power rule, its mainly just a reverse of it.

    Like if you were to take the derivative of [tex]x^2[/tex], then using the power rule it would be [tex]2x^1[/tex] or just 2x. If you were to find the antiderivative of that using [tex]\frac{x^n+1}{n+1}[/tex] (agh supposed to be x^(n+1) not (x^n)+1) then it would be [tex]x^2[/tex] again.

    If you're asking how it works, the derivative power rule derives itself from the limit of [tex]\frac{f(x+h) - f(h)}{h}[/tex] as h approaches 0. The antiderivative is a little more complicated and doesn't work in all instances (like when n=-1) so some other methods like natural logarithms need to be used.
  4. Mar 16, 2005 #3
    [tex]\frac{d}{dx}x^{n}=nx^{n-1} \Rightarrow \int{x^{n-1}dx}=\frac{x^{n}}{n}[/tex]
  5. Mar 16, 2005 #4


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    The general rule of

    [tex] (x^{z})'=z x^{z-1},z\in \mathbb{C} [/tex]

    is proven using the definition & generalized binomial formula (the one with Gamma Euler/Pochhammer symbols).

    Once u've proven the Leibniz rule & implicitely the part integration mechanism,u can use the latter to

    [tex] \int x^{z} \ dx = x^{z}\cdot x-\int (x^{z})'\cdot x \ dx =x^{z+1}-z\int x^{z} \ dx\Rightarrow \int x^{z} \ dx =\frac{1}{z+1} x^{z+1} +C ,z\neq -1



    Last edited: Mar 16, 2005
  6. Mar 16, 2005 #5
    Why is

    integral(1/x dx)=ln(x) ?

    I can't find the derivation of this in google, afgh.

  7. Mar 16, 2005 #6


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    I've proven the general case for a complex exp.other than "-1".For this singular case,i'm using the FTC which says

    [tex] \int f(x) \ dx=F(x)+C \Rightarrow \frac{dF(x)}{dx}=f(x) [/tex]

    Then i know that

    [tex] \frac{d\ln x}{dx} =\frac{1}{x} [/tex]


    [tex] \int \frac{1}{x} \ dx=\ln x+C [/tex]

  8. Mar 17, 2005 #7


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    [tex]\int_1^x \frac{1}{t}dt = \ln(x)[/tex]
    is used as the definition for [itex]\ln x[/itex].

    Otherwise, you can use implicit differentiation:

    [tex]y=\ln x \iff x = e^y \Rightarrow 1=e^y \frac{dy}{dx} \iff \frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}[/tex]

    and apply the FTC.
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