# Derivative of function $\ y=[e^{x^2}-\cot(\ln(\sqrt x+\frac 1x))]^{\sec(\frac1x)}$

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?

• There seems to errors in your formatting which may affect the function. When looking at your question, I believe you meant $e^{x^2}$ instead of $e^(x^2)$. If that is the case please fix your formatting. – Madhav Nakar Jan 23 '18 at 15:10
• Probably it was intended that you would use logarithmic differentiation for this, since $x$ appears both in the base and in the exponent. – Michael Hardy Jan 23 '18 at 15:12
• No, $e^{x^2}$ is definitely NOT equal to "2 ln(x)"! – user247327 Jan 23 '18 at 15:14
• @MichaelHardy Yeap, but as Rohan posted an answer below, it's way too big solution if i start differentiating. I think there is some shortcut in the simplifying – Stefan Todorovski Jan 23 '18 at 15:26

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.