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Math Help - Differentiating The natural Exponential Function

  1. #1
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    Differentiating The natural Exponential Function

    Differentiate the function:
    y = (x^5)(e^{-3lnx})
    Howw??
    THe answer is 2x
    Last edited by ^_^Engineer_Adam^_^; January 27th 2007 at 01:06 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Differentiate the function:
    y = (x^5)(e^{-3lnx})
    Howw??
    THe answer is 2x
    The definition of logarithms is such that:

    a=\ln(b)

    means that e^a=b, or what is the same thing e^{\ln(b)}=b, then:

    e^{-3\ln(x)}=e^{\ln(1/x^3)}=1/x^3

    so:

    y=x^2,

    hence:

    \frac{dy}{dx}=2x

    RonL
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  3. #3
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    Hello, ^_^Engineer_Adam^_^!

    I'll give it a try . . .


    Differentiate the function: . y \:= \:(x^5)(e^{-3\ln x})

    The answer is 2x

    Let e^{-3\ln x} \:=\:z

    Take logs: . \ln\left(e^{-3\ln x}\right)\:=\:\ln z\quad\Rightarrow\quad -3\ln x\cdot\ln(e)\:=\:\ln z\quad\Rightarrow\quad-3\ln x\:=\:\ln z

    . . \ln z \:=\:-3\ln x\:=\:\ln\left(x^{-3}\right)\quad\Rightarrow\quad z \:=\:x^{-3}

    Hence: . e^{-3\ln x} \:=\:x^{-3}


    The function becomes: . y \:=\:x^5\!\cdot\!x^{-3} \:=\:x^2

    Therefore: . y' \:=\:2x

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We can do it head-on . . . (not recommended).

    We have: . y \;=\;\left(x^5\right)\left(e^{-3\ln x}\right)

    Product Rule: . y' \;=\;\left(x^5\right)\left(e^{-3\ln x}\right)\left(-3\cdot\frac{1}{x}\right) + \left(5x^4\right)\left(e^{-3\ln x}\right)

    We have: . y' \;=\;-3x^4e^{-3\ln x} + 5x^4e^{-3\ln x} \;=\;2x^4e^{-3\ln x}

    Since e^{-3\ln x}\:=\:x^{-3}, we have: . y' \;=\;2x^4\!\cdot\!x^{-3}\;=\;2x

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