This is a problem from a college exam which i can't figure it out.
First i add $\ln$ on both sides: $$\ \ln y=\ln [e^{x^2}-\cot(\ln(\sqrt x+\frac 1x))]^{\sec(\frac1x)} $$
Then when i simplify it i get this: $\ \ln y=\sec(\frac1x)\ln [e^{x^2}-\cot(\ln(\sqrt x+\frac 1x))] $
EDIT: I'm not sure if i can simplify $\ \cot(\ln(\sqrt x+\frac 1x)) $ into $\ \cot(\ln(\frac{x\sqrt x+1}{x}) $ and to $\ \cot(\ln(x\sqrt x+1) - \ln(x)) $
I think this isn't simplified enough, so i can't continue with finding the derivative. How could i simplify it further?
Thanks in advance.
Basically, your function is: $$y=[e^{x^2}-\cot(\ln(\sqrt x + \frac1{x}))]^{\sec \frac1{x}} $$
Applying logarithms on both sides, we have: $$\ln y = \sec \frac1{x} \ln[e^{x^2}-\cot(\ln(\sqrt x + \frac1{x}))] $$ Taking the derivative, with respect to $x$, we get: $$y'=y\times \left((-\frac1{x^2}) \sec \frac1{x} \tan \frac1{x} \ln[e^{x^2}-\cot(\ln(\sqrt x + \frac1{x})) + \sec \frac1{x} \frac1{e^{x^2}-\cot(\ln(\sqrt x + \frac1{x}))} \times \left[2xe^{x^2} +\csc^2(\ln(\sqrt x + \frac1{x})) \times \frac1{\sqrt x + \frac1{x}}\left[\frac1{2\sqrt x} - \frac1{x^2}\right]\right]\right)$$ using successive applications of the product rule.