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...
First let's rewrite $x^{\log x} = e^{\log x\log x} = e^{\log^2 x}$ and $\sec^{3x}(x) = e^{3x\log(\sec x)}$ Then, $$ f'(x) = \sec^{3x}(x) \left(e^{\log^2 x}\right)' + e^{\log^2 x}\left(e^{3x\log(\sec x)}\right)' $$ Observe that $$ \left(e^{\log^2 x}\right)' = 2\frac{\log x}{x}e^{\log^2 x} $$ and $$ \left(e^{3x\log(\sec x)}\right)' = \left(3(\log(\sec x)) + 3x\left(\frac{1}{\sec x}\sec x \tan x\right)\right)e^{3x\log(\sec x)} $$
Can you continue from here?
EDIT: As a rule of thumb, when having variable exponents you must use the fact that $\exp\circ\log = \operatorname{Id}$ to turn them into products.
Try logarithmic differentiation, or somehow make use the fact that $a=e^{\ln a}$ for $a>0$. (you have to decide what $a$ should be).
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,
$$y'=(2\frac{\ln x}x-3\ln\cos x+3x\tan x)x^{\ln(x)}\sec(x)^{3x}.$$