Thread: x^x or 10^x

1. 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.

2. 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*}

Thanks!

4. 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 $\displaystyle y'= y(ln(x)+ 1)= x^x(ln(x)+ 1)$.