Find the derivative of $y = x^{\ln(x)}\sec(x)^{3x}$ What is the derivative of 
  $$y = x^{\ln(x)}\sec(x)^{3x}$$
I tried to find the derivative of this function but somewhere along the way I seem to have gotten lost. I started off with using the product rule and then the chain rule about 4 times and things are getting messier. Now I just don't know what to do. I would show my work here but that would be roughly 2 whole pages of calculations...

Derivatives of powers are conveniently obtained using logarithms.
$$y = x^{\ln(x)}\sec(x)^{3x}=e^{\ln^2x+3x\ln\sec x}.$$
For later simplification, we note that by the definition of the secant, $\ln\sec x=-\ln\cos x$.
Then the derivative of an exponential is the same exponential. Using the chain rule,
$$y'=(\ln^2x-3x\ln\cos x)'y.$$
The derivative of a square is twice the argument. Using the chain rule,
$$(\ln^2x)'=(\ln x)'2\ln x=2\frac{\ln x}x.$$
The derivative of a product is the sum of each factor times the derivative of the other,
$$(3x\ln\cos x)'=3\ln\cos x+3x(\ln\cos x)'.$$
And finally by the chain rule,
$$(\ln\cos x)'=-\frac{\sin x}{\cos x}=-\tan x.$$
Putting all results together,
$$y'=(2\frac{\ln x}x-3\ln\cos x+3x\tan x)x^{\ln(x)}\sec(x)^{3x}.$$
