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. **Start with the given equation:**
\[ \sin^2 x + \cos^2 y = 1 \]

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

3. **Differentiate \(\sin^2 x\):**
Using the chain rule:
\[ \frac{d}{dx}(\sin^2 x) = 2\sin x \cdot \frac{d}{dx}(\sin x) = 2\sin x \cdot \cos x \]
Therefore:
\[ \frac{d}{dx}(\sin^2 x) = 2\sin x \cos x \]

4. **Differentiate \(\cos^2 y\):**
Again, using the chain rule:
\[ \frac{d}{dx}(\cos^2 y) = 2\cos y \cdot \frac{d}{dx}(\cos y) = 2\cos y \cdot (-\sin y) \cdot \frac{dy}{dx} \]
Therefore:
\[ \frac{d}{dx}(\cos^2 y) = -2\cos y \sin y \frac{dy}{dx} \]

5. **Differentiate the right side:**
The derivative of a constant (1) is 0:
\[ \frac{d}{dx}(1) = 0 \]

6. **Combine the derivatives:**
Putting it all together, we have:
\[ 2\sin x \cos x - 2\cos y \sin y \frac{dy}{dx} = 0 \]

7. **Solve for \(\frac{dy}{dx}\):**
Rearranging the equation gives:
\[ -2\cos y \sin y \frac{dy}{dx} = -2\sin x \cos x \]
Dividing both sides by \(-2\cos y \sin y\):
\[ \frac{dy}{dx} = \frac{\sin x \cos x}{\cos y \sin y} \]

8. **Final Result:**
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\sin x \cos x}{\cos y \sin y} \]