Discussion
If \(y = \cos x\cos y\), then \(\frac{dy}{dx}\) at \(\left(\frac{\pi}{3}, \frac{\pi}{6}\right)\) is
Answer: A
The value is \(\frac{-3}{5}\).
Differentiate \(y = \cos x\cos y\) implicitly, using the product rule.
\[ \begin{aligned} y' &= -\sin x\cos y - \cos x\sin y\, y' \\ y' &= \frac{-\sin x\cos y}{1 + \cos x\sin y} \end{aligned} \]
At \(x = \frac{\pi}{3}, y = \frac{\pi}{6}\): \(\sin x\cos y = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{3}{4}\) and \(\cos x\sin y = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}\), so
\[ y' = \frac{-3/4}{1 + 1/4} = -\frac{3}{5} \]
Dropping the minus sign gives \(\frac{3}{5}\), and inverting the fraction gives \(\pm\frac{5}{3}\).
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