first derivative of f(x) = {sqrt (5x) cos [(e^-x/x)^2]}^pi?

What sort of function is that anyway? Even wolframalpha couldn't be bothered graphing it.

Anyhoo...
lets start from the inside, lets rewrite as f=(g.h)^π where g=sqrt (5x) and h=cos [e^(-2x)/x^2]
f'=π(g.h)^(π-1)(g'h+h'g)
here g'=(sqrt(5)/(2sqrt(x))
and if we think of h=cos(m) where m=e^(-2x)/x^2
then h'=-m'sin(m)
use the quotient rule to find m'
m'=[(2x^2*e^(-2x)+2xe^(-2x))/x^4]
so putting it all together we ought to get
f'=π(g.cos(m))^(π-1)*..
(g'cos(m)-m'sin(m)g)
should be
=π{sqrt (5x) cos [e^(-2x)/x^2]}^(π-1)*..
(sqrt(5)/(2sqrt(x))cos(e^(-2x)/x^2)+..
(sqrt(5x))*[(2x^2*e^(-2x)+..
.2xe^(-2x))/x^4]sin(e^(-2x)/x^2)

which can probably be simplified somewhat but stuff it, if they're going to give you a gross question, give em a gross answer.

Also sorry about the line breaks, it wasn't displaying the whole line.