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Math Help - Logarithmic Diff.....

  1. #1
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    Logarithmic Diff.....

    Find the derivative of the function
    f(x) = x^{\sqrt{x}}; x>0
    Do i have to use logarithmic differentiation?
    The answer according to the back of book is....
    x^{\sqrt{x} - \frac{1}{2}}(1 + \frac{1}{2} \ln x)
    how?
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the derivative of the function
    f(x) = x^{\sqrt{x}}; x>0
    Do i have to use logarithmic differentiation?
    The answer according to the back of book is....
    x^{\sqrt{x} - \frac{1}{2}}(1 + \frac{1}{2} \ln x)
    how?
    You can write,
    x=e^{\ln x}
    Thus,
    x^{\sqrt{x}}=e^{\ln x\sqrt{x}}
    Find the derivative of the inside function,
    u=\ln x\sqrt{x}
    u'=(\ln x)'\sqrt{x}+(\ln x)(\sqrt{x})'
    u'=\frac{1}{x}\cdot \sqrt{x} + \ln x\cdot \frac{1}{2\sqrt{x}}
    Irrationalize the denominator,
    u'=\frac{1}{\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}
    u'=\frac{2+\ln x}{2\sqrt{x}}=\left( \frac{2+\ln x}{2} \right) \frac{1}{\sqrt{x}}=(1+\frac{1}{2}\ln x)x^{-1/2}
    Now times the outer function,
    (1+\frac{1}{2}\ln x)x^{-1/2}e^{\ln x\sqrt{x}}=(1+\frac{1}{2}\ln x)x^{-1/2}x^{\sqrt{x}}
    Laws of exponents,
    (1+\frac{1}{2}\ln x)x^{\sqrt{x} - \frac{1}{2}}
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the derivative of the function
    f(x) = x^{\sqrt{x}}; x>0
    Do i have to use logarithmic differentiation?
    The answer according to the back of book is....
    x^{\sqrt{x} - \frac{1}{2}}(1 + \frac{1}{2} \ln x)
    how?
    \ln(f(x))=\sqrt{x}\ln(x)

    so:

    \frac{1}{f(x)}\,\frac{df}{dx}=\frac{1}{2\sqrt{x}}\  ln(x)+\frac{\sqrt{x}}{x}

    so:

    <br />
\frac{df}{dx}= \frac{x^{\sqrt{x}} \ln(x)}{2\sqrt{x}}+\frac{x^{\sqrt{x}}\sqrt{x}}{x}<br />

    which when all the powers of x are collected together gives the required result:

    <br />
\frac{df}{dx}= \frac{x^{\sqrt{x}-1/2} \ln(x)}{2}+x^{\sqrt{x}-1/2}<br />

    or:

    \frac{df}{dx}=x^{\sqrt{x} - \frac{1}{2}}(1 + \frac{1}{2} \ln x)
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  4. #4
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the derivative of the function
    f(x) = x^{\sqrt{x}}; x>0
    Do i have to use logarithmic differentiation?
    The answer according to the back of book is....
    x^{\sqrt{x} - \frac{1}{2}}(1 + \frac{1}{2} \ln x)
    how?
    Here is one way.

    f(x) = x^(sqrt(x))

    Let y = f(x)
    So,
    y = x^(sqrt(x))
    Take the ln of both sides,
    ln(y) = sqrt(x) *ln(x)
    Differentiate both sides with respect to x,
    (1/y)y' = [sqrt(x) *1/x] +[ln(x) *(1/2)x^(-1/2)]
    (1/y)y' = [1/sqrt(x)] +[1/(2sqrt(x)) *ln(x)]
    Factor the RHS by 1/sqrt(x),
    (1/y)y' = (1/sqrt(x))[1 +(1/2)ln(x)]
    Multiply both sides by y,
    And since 1/sqrt(x) = x^(-1/2),
    y' = (x^sqrt(x))*(x^(-1/2))[1 +(1/2)ln(x)]
    y' = [x^(sqrt(x) -(1/2)][1 +(1/2)ln(x)] -----answer at the back of the book.
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