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$$\dfrac {dx}{dy} = 2\sec 2y \tan 2y = 2x \tan 2y$$

$$=> \tan^2 2y = sec^2 2y -1$$ $$=> \tan 2y = \pm \sqrt {x^2 -1}$$ $$=> \dfrac {dy}{dx} = \pm \dfrac 1 {2x\sqrt{x^2 -1}}$$

But the solution paper specifically states $$=> \dfrac {dy}{dx} = \dfrac 1 {2x\sqrt{x^2 -1}}$$

There were no range given in the question. So why drop negative differential?

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    $\begingroup$ To me, you are correct. $\endgroup$ – Claude Leibovici Jun 11 '15 at 8:36
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The writer was probably thinking of $v=2y$ being in the standard range of arcsecant, which is $0\le v\le \pi,\ v\ne \frac{\pi}2$. Using that range, the derivative of arcsecant is indeed always positive, so the plus-or-minus sign can be dropped. Indeed, the standard derivative formula is

$$\frac{d}{dx}\sec^{-1}u=\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx},\quad |u|>1$$

However, your problem is stated in terms of secant, not arcsecant, and no limits are placed on $y$. Therefore you are correct: there must be a plus-or-minus sign in the answer.

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