Question

Help figuring out the formula for \[\frac{d}{dx}(x^x)\].

I think I'm close.

Answer 1

\[x^x=\exp(x\ln(x))\] then the chain rule and product rule

Answer 2

We can write y=x^x = e^{x lnx} and then differentiate using the chain rule.

Answer 3

i think this one is a happy accident as well

Answer 4

power rule + exponential rule , or is it the product of the 2?

Answer 5

\[D_x[x^x]=x~x^{x-1} = x^x\]\[D_x[x^x]=x^x ln(x)\]
\[x^x+x^x~ln(x)=x^x(ln(x)+1)\]
happy accident :)

Answer 6

ahh, those are both better methods than what I was trying.

\[y = a^x\]
\[\log {a}{y} = x\]
\[ln y = \frac{x}{log a e}\]
\[\frac{dy}{y} = \frac{dx}{log a e}\]
\[\frac{dy}{dx} = \frac{a^x}{log a e}\]

Would it also work with this line of thinking?

Sorry, I can't figure out how to get log base a of e to work in $\LaTeX$

Answer 7

@amistre64 lol

Answer 8

@datanewb if you do it that way, it should be
\[y=x^x\]
\[\ln(y)=x\ln(x)\]

Answer 9

all the work is still the same
the original function times the derivative if \(x\ln(x)\) is your final answer

Answer 10

ahh, thank you! I see now... I just need to review log operations. Thank you.