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Thread: x^x or 10^x

  1. #1
    Senior Member Paze's Avatar
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    x^x or 10^x

    Hi MHF.

    Is there a difference between x^x and, say, 10^x when differentiating?

    When I differentiate x^x I get x^x(lnx+1) but when I differentiate 10^x I seem to get 10^x(log10)

    How is a number^x differentiated? I understand the x^x part.
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  2. #2
    MHF Contributor
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    Re: x^x or 10^x

    $\displaystyle \displaystyle \begin{align*} y &= 10^x \\ &= e^{\ln{ \left( 10^x \right) } } \\ &= e^{x\ln{(10)}} \end{align*}$

    Now let $\displaystyle \displaystyle \begin{align*} u = x\ln{(10)} \implies y = e^u \end{align*}$, then $\displaystyle \displaystyle \begin{align*} \frac{du}{dx} = \ln{(10)} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \frac{dy}{du} = e^u = e^{ x\ln{(10)}} = e^{\ln{ \left( 10^x \right) } } = 10^x \end{align*}$, so $\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = 10^x\ln{(10)} \end{align*}$.

    As for $\displaystyle \displaystyle \begin{align*} x^x \end{align*}$, first note that it is only continuous, and so only differentiable, when $\displaystyle \displaystyle \begin{align*} x \geq 0 \end{align*}$, then

    $\displaystyle \displaystyle \begin{align*} y &= x^x \\ \ln{(y)} &= \ln{ \left( x^x \right) } \\ \ln{(y)} &= x\ln{(x)} \\ \frac{d}{dx} \left[ \ln{(y)} \right] &= \frac{d}{dx} \left[ x\ln{(x)} \right] \\ &= \frac{1}{y}\,\frac{dy}{dx} &= \ln{(x)} + 1 \\ \frac{dy}{dx} &= y \left[ \ln{(x)} + 1 \right] \\ \frac{dy}{dx} &= x^x \left[ \ln{(x)} + 1 \right] \end{align*}$
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  3. #3
    Senior Member Paze's Avatar
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    Re: x^x or 10^x

    Thanks!
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  4. #4
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    Re: x^x or 10^x

    Equivalently, if $\displaystyle y= 10^x$ then $\displaystyle ln(y)= ln(10^x)= xln(10)$. Differentiating, $\displaystyle \frac{1}{y}y'= ln(10)$ so that $\displaystyle y'= ln(10)y= [ln(10)]10^x$.
    Similarly, if $\displaystyle y= x^x$ then $\displaystyle ln(y)= x ln(x)$. Differentiating (using the product rule on the right), $\displaystyle \frac{1}{y}y'= ln(x)+ x(1/x)= ln(x)+ 1$ so that $\displaystyle y'= y(ln(x)+ 1)= x^x(ln(x)+ 1)$.
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