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Math Help - Derivatives

  1. #1
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    Derivatives

    How do you do these two derivatives? (Work below.)

    1. y = e^{(1+lnx)}

    2. y = x^{lnx}





    ------------
    ------------
    1. y = e^{(1+lnx)}

    y' = \frac{1}{x}(e^{(1+lnx)})

    Actual answer:  y' = e

    --------------------

    2. y = x^{lnx}

    y' = \frac{1}{x}(lnx)(x^{lnx})

    Actual answer:
    y' = \frac{2 (lnx)( x^{lnx})}{x}
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Cursed View Post
    How do you do these two derivatives? (Work below.)

    1. y = e^{(1+lnx)}

    2. y = x^{lnx}





    ------------
    ------------
    1. y = e^{(1+lnx)}

    y' = \frac{1}{x}(e^{(1+lnx)})

    Actual answer:  y' = e

    --------------------
    note that this is the answer you have!

    recall, e^{\ln x} = x. you can use this fact to simplify your answer to get what you have stated.

    or you could have done it from the beginning: e^{1 + \ln x} = e \cdot e^{\ln x} = ex ...

    2. y = x^{lnx}

    y' = \frac{1}{x}(lnx)(x^{lnx})

    Actual answer:
    y' = \frac{2 (lnx)( x^{lnx})}{x}
    hint: note that x^{\ln x} = e^{\ln x^{\ln x}} = e^{\ln x \cdot \ln x} = e^{(\ln x)^2}
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  3. #3
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    Hello, Cursed!

    The first one is a trick question . . .


    1)\;\;y \:=\: e^{1+\ln x}

    We have: . y \;=\;e^{1+\ln x} \;=\;e^1\cdot e^{\ln x} \;=\;e\cdot x

    Therefore: .  y \:=\:ex \quad\Rightarrow\quad y\:\!' \:=\:e




    2)\;\;y \:=\: x^{\ln x}

    This requires Logarithmic Differentiation.


    Take logs: . \ln(y) \;=\;\ln\left(x^{\ln x}\right) \;=\;\ln x\cdot\ln x  \quad\Rightarrow\quad \ln(y) \;=\;\left[\ln x\right]^2


    Differentiate implicitly: . \frac{1}{y}\cdot y\:\!' \;=\;2\cdot\ln x \cdot\frac{1}{x} \quad\Rightarrow\quad \frac{y\:\!'}{y} \;=\;\frac{2\ln x}{x}

    Then: . y' \;=\;y\cdot\frac{2\ln x}{x} \;=\;\frac{2\,x^{\ln x}\ln x}{x}

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