Integration: derive y = (x^sinx)^(x^sinx)

[johnboy]

Derive y= (x^sinx)^(x^sinx)

I'm not sure if I did it correctly, but I got that the derivative is:

dy/dx = (x^sinx)^(x^sinx) * (x^sinx)(cosx/x+lncosx)(ln(x^sinx) + 1/ (cosx/x + ln cosx)

I didn't simplify it yet, but does it look right so far? Thanks.

 

[skeeter]

preliminary calculation ...

let \(\displaystyle \L u = x^{sinx}\)

\(\displaystyle \L ln(u) = sinx ln(x)\)

\(\displaystyle \L \frac{u'}{u} = \frac{sinx}{x} + cosx ln(x)\)

\(\displaystyle \L u' = x^{sinx}[\frac{sinx}{x} + cosx ln(x)]\)

now, back to the original problem ...

\(\displaystyle \L y = u^u\)

\(\displaystyle \L ln(y) = u ln(u)\)

\(\displaystyle \L \frac{d}{dx}[ln(y) = u ln(u)]\)

\(\displaystyle \L \frac{y'}{y} = u \frac{u'}{u} + u' ln(u)\)

\(\displaystyle \L \frac{y'}{y} = u'[1 + ln(u)]\)

\(\displaystyle \L y' = u^u u'[1 + ln(u)]\)

\(\displaystyle \L y' = (x^{sinx})^{x^{sinx}+1}[\frac{sinx}{x} + cosx ln(x)][1 + sinx ln(x)]\)

lots of places for a possible mistake ... hopefully someone else will check it out.

 

[pka]

Skeeter, I think that you have done it!
But what a sadistic problem: what in the world does it teach?
There are so many similar and simpler problems which do the same thing.