My work is as follows. Criticism welcomed.

$$y = \cot^2(\sin\theta) = (\cot(\sin\theta))^2$$

Power Rule combined with the Chain Rule:

$$\begin{align} y' & = 2(\cot(\sin \theta)) \cdot \frac d{dx}(\cot(\sin\theta)) \cdot \frac d{dx}(\sin \theta) \\ \\ &= 2(\cot(\sin \theta)) \cdot (- \csc^2(\sin \theta)) \cdot \cos \theta \end{align}$$

  • 4
    $\begingroup$ That's correct, Cetshwayo. $\endgroup$ – Namaste Mar 15 '15 at 13:46
  • $\begingroup$ Is there a way I can make my answer more elegant? $\endgroup$ – Cetshwayo Mar 15 '15 at 13:51
  • 3
    $\begingroup$ you can just shift the minus sign to the leftmost side? $\endgroup$ – RE60K Mar 15 '15 at 13:52

As I said in my comment above: Your work is correct.

As far as making it "elegant", I would simply pull the negative (the coefficient of $\csc^2(\sin\theta))$ to the front: $$-2\cot(\sin\theta)\csc^2(\sin\theta)(\cos \theta),$$

Other than that, you might want to bring the factor of $\cos \theta$ to the front as well: $$-2(\cos \theta) \cot(\sin \theta)\csc^2(\sin\theta).$$

Otherwise, elegance is in the eye of the beholder!

  • 1
    $\begingroup$ You're welcome, Cetshwayo! $\endgroup$ – Namaste Mar 15 '15 at 14:52

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