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Math Help - derivative of a function

  1. #1
    Boo
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    derivative of a function

    Please, what woudl be the derivative of a function?
    I foudn sth like:

    (f(x)=x^a
    (f(x)'=x^a(a)'
    is that ok???
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  2. #2
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    Re: derivative of a function

    Quote Originally Posted by Boo View Post
    Please, what woudl be the derivative of a function?
    I foudn sth like:
    (f(x)=x^a find (f(x)'=x^a(a)'

    If a is a non-zero constant then

    \left(x^a\right)^{\prime}=a\left(x\right)^{a-1}
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  3. #3
    Boo
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    Re: derivative of a function

    Hello!
    I ment, if a = function!
    some tol dme it is different!!!
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: derivative of a function

    Yep, that is different.

    (x^{a(x)})'

    = (e^{a(x) \ln x})'

    = e^{a(x) \ln x} \cdot (a'(x) \ln x + a(x) \cdot \frac 1 x)

    = x^{a(x)} \cdot (a'(x) \ln x + a(x) \cdot \frac 1 x)

    = x^{a(x)} a'(x) \ln x + a(x) x^{a(x) - 1}
    Thanks from Boo
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    Boo
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    Re: derivative of a function

    Yes, that is exactly what i ment!!
    Do we "add" lnx in that formula or is that a typo? Many thans! I ment, is the original
    x^aor x^{alnx}
    Last edited by Boo; February 4th 2013 at 11:34 PM.
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    Re: derivative of a function

    Quote Originally Posted by Boo View Post
    Please, what woudl be the derivative of a function?
    I foudn sth like:

    (f(x)=x^a
    (f(x)'=x^a(a)'
    is that ok???
    Logarithmic differentiation works well here...

    \displaystyle \begin{align*} y &= x^{a} \\ \ln{(y)} &= \ln{\left( x^a \right)} \\ \ln{(y)} &= a\ln{(x)} \\ \frac{d}{dx} \left[ \ln{(y)} \right] &= \frac{d}{dx} \left[ a\ln{(x)} \right] \\ \frac{d}{dy}\left[ \ln{(y)} \right] \frac{dy}{dx} &= a\,\frac{d}{dx} \left[ \ln{(x)} \right] + \ln{(x)} \,\frac{d}{dx} \left( a \right) \\ \frac{1}{y} \, \frac{dy}{dx} &= a \left( \frac{1}{x} \right) + \ln{(x)}\,\frac{da}{dx} \\ \frac{dy}{dx} &= y \left[ \frac{a}{x} + \ln{(x)} \,\frac{da}{dx} \right] \\ \frac{dy}{dx} &= x^a \left[ \frac{a}{x} + \ln{(x)} \, \frac{da}{dx} \right]  \end{align*}
    Thanks from Boo
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    Super Member ILikeSerena's Avatar
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    Re: derivative of a function

    Quote Originally Posted by Boo View Post
    Yes, that is exactly what i ment!!
    Do we "add" lnx in that formula or is that a typo? Many thans! I ment, is the original
    x^aor x^{alnx}
    There is no typo. lnx is in the formula, but it is not added but multiplied.
    I guess I could also write it as:

    (x^a)' = x^a a' \ln x + ax^{a-1}
    Thanks from Boo
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  8. #8
    Boo
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    Re: derivative of a function

    got it!!!
    many many thanks!!!
    Last edited by Boo; February 5th 2013 at 01:35 AM.
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  9. #9
    Super Member ILikeSerena's Avatar
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    Re: derivative of a function

    Quote Originally Posted by Boo View Post
    Yes, I see!
    How did U get the first line, sorry, I really forgot...
    (x^a(x))'=(e^{a(x)lnx}
    Many thanks¨!!!
    It follows from the calculation rules for logarithm and exponentials.

    Suppose q=\ln p, then:
    p = e^q by definition.
    If we substitute q, we get:
    [1] p = e^{\ln p}

    We also have the rule:
    [2] \ln p^q = q \ln p

    So to make that first step more explicit.
    Using [1] we have:
    x^a = e^{\ln(x^a)}
    And using [2] we have:

    e^{\ln(x^a)} = e^{a \ln x}
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