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Math Help - differentiation help

  1. #1
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    differentiation help

    Hi
    I need to find the derivative of the following
    y=(4sqrtx)^lnx
    Ummmm, so I'm not really sure how to type that. It is the ("forth root" of x) to the power of (lnx).
    I thought that maybe you could change it to
    y=x^(.25lnx)
    lny=ln.25lnxlnx
    lny=ln.25(lnx)^2
    (1/y)(y')=f'g+fg'
    (1/y)(y')=(0(lnx))+(ln.25)(2(lnx)(1/x))
    y'=[(2(ln.25)(lnx))/x]((4sqrtx)^lnx)
    Is this assumption incorrect?
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  2. #2
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    Quote Originally Posted by sberxa View Post
    Hi
    I need to find the derivative of the following
    y=(4sqrtx)^lnx
    Ummmm, so I'm not really sure how to type that. It is the ("forth root" of x) to the power of (lnx).
    I thought that maybe you could change it to
    y=x^(.25lnx)
    lny=ln.25lnxlnx
    lny=ln.25(lnx)^2
    (1/y)(y')=f'g+fg'
    (1/y)(y')=(0(lnx))+(ln.25)(2(lnx)(1/x))
    y'=[(2(ln.25)(lnx))/x]((4sqrtx)^lnx)
    Is this assumption incorrect?
    y = (\sqrt[4]{x})^{\ln{x}}

    y = \left(x^{\frac{1}{4}}\right)^{\ln{x}}

    y = x^{\frac{1}{4}\ln{x}}

    \ln{y} = \ln{x^{\frac{1}{4}\ln{x}}}

    \ln{y} = \frac{1}{4}\ln{x}\ln{x}

    \ln{y} = \frac{1}{4}(\ln{x})^2

    \ln{y} = \left(\frac{1}{2}\ln{x}\right)^2

    y = e^{\left(\frac{1}{2}\ln{x}\right)^2}.


    Now use the Chain rule.

    You should get

    \frac{dy}{dx} = \frac{\ln{x}e^{(\frac{1}{2}\ln{x})^2}}{2x}.
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  3. #3
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    \ln{y} = \frac{1}{4}(\ln{x})^2

     \frac{y'}{y} = \frac{1}{2}(\ln{x}) \frac{1}{x}

    y' = \frac{y \ln{x}}{2x} = \frac{(\sqrt[4]{x})^{\ln{x}} \ln{x}}{2x}
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  4. #4
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    Quote Originally Posted by Prove It View Post
    y = (\sqrt[4]{x})^{\ln{x}}

    y = \left(x^{\frac{1}{4}}\right)^{\ln{x}}

    y = x^{\frac{1}{4}\ln{x}}

    \ln{y} = \ln{x^{\frac{1}{4}\ln{x}}}

    \ln{y} = \frac{1}{4}\ln{x}\ln{x}

    \ln{y} = \frac{1}{4}(\ln{x})^2

    \ln{y} = \left(\frac{1}{2}\ln{x}\right)^2

    y = e^{\left(\frac{1}{2}\ln{x}\right)^2}.


    Now use the Chain rule.

    You should get

    \frac{dy}{dx} = \frac{\ln{x}e^{(\frac{1}{2}\ln{x})^2}}{2x}.
    Another way to do all that:
    y = (\sqrt[4]{x})^{\ln{x}}

    y = e^{\ln{(\sqrt[4]{x})^{\ln{x}}}}

    y = e^{(\frac{1}{2}\ln{x})^2}
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