Question

If $y=\cos x \cos y$, then $\frac{dy}{dx}$ at $\left(\frac{\pi}{3},\frac{\pi}{6}\right)$ is:

• A. $\frac{-3}{5}$
• B. $\frac{3}{5}$
• C. $\frac{-5}{3}$
• D. $\frac{-4}{3}$
• E. $\frac{-8}{3}$

Hint:

Group all terms containing $\frac{dy}{dx}$ on one side of the equation to isolate the derivative.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given an implicit equation relating x and y. We need to find the derivative \( \frac{dy}{dx} \) at a specific point. This requires implicit differentiation.
Step 2: Key Formula or Approach:
We will differentiate both sides of the equation with respect to x, remembering to apply the chain rule for terms involving y. The derivative of y with respect to x is \( \frac{dy}{dx} \). We will also need the product rule.
Product Rule: \( (uv)' = u'v + uv' \).
Step 3: Detailed Explanation:
The given equation is \( y = \cos x \cos y \).
Differentiate both sides with respect to x:
\[ \frac{d}{dx}(y) = \frac{d}{dx}(\cos x \cos y) \] The left side is straightforward:
\[ \frac{dy}{dx} \] For the right side, we use the product rule with \( u = \cos x \) and \( v = \cos y \).
\( u' = \frac{d}{dx}(\cos x) = -\sin x \)
\( v' = \frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx} \) (by the chain rule)
Applying the product rule \( u'v + uv' \):
\[ \frac{d}{dx}(\cos x \cos y) = (-\sin x)(\cos y) + (\cos x)(-\sin y \frac{dy}{dx}) \] \[ = -\sin x \cos y - \cos x \sin y \frac{dy}{dx} \] Now, set the differentiated left and right sides equal:
\[ \frac{dy}{dx} = -\sin x \cos y - \cos x \sin y \frac{dy}{dx} \] Our goal is to solve for \( \frac{dy}{dx} \). Group all terms with \( \frac{dy}{dx} \) on one side:
\[ \frac{dy}{dx} + \cos x \sin y \frac{dy}{dx} = -\sin x \cos y \] Factor out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx}(1 + \cos x \sin y) = -\sin x \cos y \] Isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{-\sin x \cos y}{1 + \cos x \sin y} \] Now, we need to evaluate this derivative at the point \( (x, y) = (\frac{\pi}{3}, \frac{\pi}{6}) \).
We need the values:
- \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \)
- \( \cos(\frac{\pi}{3}) = \frac{1}{2} \)
- \( \sin(\frac{\pi}{6}) = \frac{1}{2} \)
- \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \)
Substitute these into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{-(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})}{1 + (\frac{1}{2})(\frac{1}{2})} = \frac{-\frac{3}{4}}{1 + \frac{1}{4}} = \frac{-\frac{3}{4}}{\frac{5}{4}} \] \[ = -\frac{3}{4} \cdot \frac{4}{5} = -\frac{3}{5} \] This matches the provided answer. Step 4: Final Answer:
The value of \( \frac{dy}{dx} \) at the given point is \( -\frac{3}{5} \).