Question

could you guys help me to find the derivative of this function.

Answer 1

Logarithms. Lots of logarithms.

Answer 2

First compute some simpler derivatives:
\[\large \frac{d}{dx}x^a=ax^{x-1}\]
\[\large \frac{d}{dx}a^x=\frac{d}{dx}e^{\ln a^x}=\frac{d}{dx}e^{x\ln a}=(\ln a)e^{x\ln a}=(\ln a)a^x\]
\[\large\frac{d}{dx} x^x=\frac{d}{dx}e^{\ln x^x}=\frac{d}{dx}e^{x\ln x}=(\ln x+1)e^{x\ln x}=(\ln x+1)x^x\]

Answer 3

From here it's a matter of applying the chain rule.
\[\large\frac{d}{dx}a^{x^x}=\frac{d}{dx}e^{\ln a^{x^x}}=\frac{d}{dx}e^{x^x\ln a}=\ln a(\ln x+1)x^xe^{x^x\ln a}\\
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad~~~\large=\ln a(\ln x+1)x^x~a^{x^x}\]
Try the other terms.

Answer 4

\[\frac{ d }{ dx }[f(x)+g(x)+h(x)+...]=\frac{ d }{ dx }f(x)+\frac{ d }{ dx }g(x)+\frac{ d }{ dx }h(x)+...\]Differentiate each term with respect to x and then write the respective derivatives as a sum. @SithsAndGiggles you're supposed to teach, not give out answers.

Answer 5

I don't have a license to teach

Answer 6

oh thanks

Answer 7

ill try to work it out myself will come back when i have something :D