`If sin^(2) x + cos^(2) y = 1 , " then " (dy)/(dx)` is equal to

Options

A. `(sin 2x)/(sin 2y)`
B. `(sin^(2)y)/(sin 2x)`
C. `(sin^(2) x)/(sin^(2)y)`
D. `(sin^(2) y)/(sin^(2) x)`

Explanation

To solve the problem where \( \sin^2 x + \cos^2 y = 1 \) and we need to find \( \frac{dy}{dx} \), we will use implicit differentiation. Here’s the step-by-step solution:

### Step 1: Differentiate both sides with respect to \( x \)

We start with the equation:
\[ \sin^2 x + \cos^2 y = 1 \]

Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^2 y) = \frac{d}{dx}(1) \]

### Step 2: Apply the chain rule

Using the chain rule, we differentiate \( \sin^2 x \) and \( \cos^2 y \):
\[ \frac{d}{dx}(\sin^2 x) = 2 \sin x \cdot \cos x \]
\[ \frac{d}{dx}(\cos^2 y) = 2 \cos y \cdot (-\sin y) \cdot \frac{dy}{dx} \]

So we have:
\[ 2 \sin x \cos x - 2 \cos y \sin y \frac{dy}{dx} = 0 \]

### Step 3: Rearrange the equation

Now, we can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ 2 \sin x \cos x = 2 \cos y \sin y \frac{dy}{dx} \]

Dividing both sides by \( 2 \cos y \sin y \):
\[ \frac{dy}{dx} = \frac{\sin x \cos x}{\cos y \sin y} \]

### Step 4: Simplify using double angle identities

Using the double angle identities:
\[ \sin 2x = 2 \sin x \cos x \quad \text{and} \quad \sin 2y = 2 \sin y \cos y \]

We can rewrite the expression:
\[ \frac{dy}{dx} = \frac{\frac{1}{2} \sin 2x}{\frac{1}{2} \sin 2y} = \frac{\sin 2x}{\sin 2y} \]

### Final Answer

Thus, the final result is:
\[ \frac{dy}{dx} = \frac{\sin 2x}{\sin 2y} \]

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