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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Madhav Nakar Jan 23 '18 at 15:10
  • $\begingroup$ Probably it was intended that you would use logarithmic differentiation for this, since $x$ appears both in the base and in the exponent. $\endgroup$ – Michael Hardy Jan 23 '18 at 15:12
  • $\begingroup$ No, $e^{x^2}$ is definitely NOT equal to "2 ln(x)"! $\endgroup$ – user247327 Jan 23 '18 at 15:14
  • $\begingroup$ @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 $\endgroup$ – 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.

  • $\begingroup$ That's far too complicated for a college exam, so something must be wrong (I don't mean that your answer is incorrect). $\endgroup$ – Weijun Zhou Jan 23 '18 at 15:21
  • $\begingroup$ I thought the same... but there must be another solution (shorter) since the answer is too big. I think there should be more simplifying before calculating the derivative $\endgroup$ – Stefan Todorovski Jan 23 '18 at 15:24
  • $\begingroup$ @WeijunZhou Don't underestimate college. Anything can happen in an exam. $\endgroup$ – Rohan Jan 23 '18 at 16:06

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