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Thread: derivativ

  1. #1
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    derivativ

    find f prim (1) when f(x)= x^1+sqrt x .

    thanks
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  2. #2
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    Hi

    $\displaystyle f(x) = x^{1+\sqrt{x}} = e^{(1+\sqrt{x})ln(x)}$

    The first derivative of

    $\displaystyle e^{u(x)}$ is $\displaystyle u'(x) e^{u(x)}$

    Here $\displaystyle u(x) = (1+\sqrt{x})ln(x)$

    u(x) is a product of 2 functions
    $\displaystyle u'(x) = \frac{1}{2 \sqrt{x}}ln(x) + \frac{1+\sqrt{x}}{x}$
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  3. #3
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    hi !
    I dont know this is power fuction or exponetial function can you explain more .
    thank you
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  4. #4
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    familiar with logarithmic differentiation?

    $\displaystyle y = x^{1+\sqrt{x}}$

    $\displaystyle \ln{y} = \ln\left(x^{1+\sqrt{x}}\right)$

    $\displaystyle \ln{y} = (1 + \sqrt{x})\ln{x}
    $

    $\displaystyle \frac{d}{dx}[\ln{y} = (1 + \sqrt{x})\ln{x}]$

    $\displaystyle \frac{y'}{y} = (1 + \sqrt{x})\cdot \frac{1}{x} + \ln{x} \cdot \frac{1}{2\sqrt{x}}$

    $\displaystyle y' = y\left((1 + \sqrt{x})\cdot \frac{1}{x} + \ln{x} \cdot \frac{1}{2\sqrt{x}}\right)$

    $\displaystyle y' = x^{1+\sqrt{x}}\left((1 + \sqrt{x})\cdot \frac{1}{x} + \ln{x} \cdot \frac{1}{2\sqrt{x}}\right)$
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