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Thread: Derivative of a fractional exponent proof?

  1. #1
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    Derivative of a fractional exponent proof?

    such that if
    $\displaystyle f(x) = x^{p/q}$ then $\displaystyle f'(x)={p/q}x^{p/q-1}$

    using the definition of a derivative (the 'lim as h tends to zero' one), the formula $\displaystyle a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-1}b+...+b^{n-1})$
    and fact that the derivative of $\displaystyle f(x)=x^{1/q}$ is $\displaystyle 1/qx^{1/q-1}$ (proved earlier using the above formula, where the "h" falls out nicely)

    I have achieved the result implicitly, but did not use the def-n of the derivative or the $\displaystyle a^{n}-b^{n}$ formula
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  2. #2
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    You can prove that $\displaystyle \displaystyle \frac{d}{dx}(x^n) = n\,x^{n-1}$ for all $\displaystyle \displaystyle n$ using the chain rule.


    $\displaystyle \displaystyle y = x^n$

    $\displaystyle \displaystyle = e^{\ln{(x^n)}}$

    $\displaystyle \displaystyle = e^{n\ln{x}}$.


    Now let $\displaystyle \displaystyle u = n\ln{x}$ so that $\displaystyle \displaystyle y = e^u$.


    $\displaystyle \displaystyle \frac{du}{dx} = \frac{n}{x}$.


    $\displaystyle \displaystyle \frac{dy}{du} = e^u = e^{n\ln{x}}$.


    Thus $\displaystyle \displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$

    $\displaystyle \displaystyle = \frac{n\,e^{n\ln{x}}}{x}$

    $\displaystyle \displaystyle = \frac{n\,e^{\ln{(x^n)}}}{x}$

    $\displaystyle \displaystyle = \frac{n\,x^n}{x}$

    $\displaystyle \displaystyle = n\,x^{n-1}$.

    Q.E.D.


    Of course, you will need to have proved the chain rule and the derivatives of $\displaystyle \displaystyle e^x$ and $\displaystyle \displaystyle \ln{x}$ beforehand...
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  3. #3
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    thank you for this, but i am required to construct a proof from first principles and only using the information given
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