Find `(dy)/(dx)` If `sin^2 x +cos^2 y=1`

Answer

To find \(\frac{dy}{dx}\) for the equation \( \sin^2 x + \cos^2 y = 1 \), we will differentiate both sides of the equation with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate both sides of the equation**:
\[ \frac{d}{dx}(\sin^2 x + \cos^2 y) = \frac{d}{dx}(1) \]
Since the derivative of a constant (1) is 0, we have:
\[ \frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^2 y) = 0 \]

2. **Apply the chain rule to differentiate \(\sin^2 x\)**:
\[ \frac{d}{dx}(\sin^2 x) = 2\sin x \cdot \frac{d}{dx}(\sin x) = 2\sin x \cdot \cos x \]
Thus, we have:
\[ 2\sin x \cos x + \frac{d}{dx}(\cos^2 y) = 0 \]

3. **Apply the chain rule to differentiate \(\cos^2 y\)**:
\[ \frac{d}{dx}(\cos^2 y) = 2\cos y \cdot \frac{d}{dx}(\cos y) = 2\cos y \cdot (-\sin y \cdot \frac{dy}{dx}) \]
Therefore, substituting this back, we get:
\[ 2\sin x \cos x - 2\cos y \sin y \cdot \frac{dy}{dx} = 0 \]

4. **Rearranging the equation**:
\[ 2\sin x \cos x = 2\cos y \sin y \cdot \frac{dy}{dx} \]

5. **Isolate \(\frac{dy}{dx}\)**:
\[ \frac{dy}{dx} = \frac{2\sin x \cos x}{2\cos y \sin y} \]
Simplifying this gives:
\[ \frac{dy}{dx} = \frac{\sin x \cos x}{\cos y \sin y} \]

6. **Using the double angle identity**:
We can express \(\sin x \cos x\) as \(\frac{1}{2}\sin(2x)\) and \(\sin y \cos y\) as \(\frac{1}{2}\sin(2y)\):
\[ \frac{dy}{dx} = \frac{\frac{1}{2}\sin(2x)}{\frac{1}{2}\sin(2y)} = \frac{\sin(2x)}{\sin(2y)} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{\sin(2x)}{\sin(2y)} \]