Discussion
If \(u = sec^{-1}(-sec2\theta)\) and \(v = cos\,\theta\) , then \(\frac{du}{dv}\) at \(\theta = \frac{\pi}{4}\) ,is equal to
Answer: B
The value is \(2\sqrt{2}\).
Simplify u first. Using \(\sec^{-1}(-y) = \pi - \sec^{-1}y\) and \(\sec^{-1}(\sec 2\theta) = 2\theta\) for \(2\theta\) in the principal range,
\[ \begin{aligned} u = \pi - 2\theta &\Rightarrow \frac{du}{d\theta} = -2 \\ v = \cos\theta &\Rightarrow \frac{dv}{d\theta} = -\sin\theta \end{aligned} \]
\[ \begin{aligned} \frac{du}{dv} &= \frac{-2}{-\sin\theta} = \frac{2}{\sin\theta} \\ &= \frac{2}{1/\sqrt{2}} = 2\sqrt{2} \end{aligned} \]
(At exactly \(\theta = \frac{\pi}{4}\), \(\sec 2\theta\) is undefined; the value is that of the derivative formula, which is the same on both sides of that point.) \(\sqrt{2}\) loses the factor 2 from \(2\theta\), and \(-\sqrt{2}\) also loses a sign.
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