find dy/dx of sin^2(x)+cos^2(y)=1 without using log

To find \(\frac{dy}{dx}\) of the given equation \(\sin^2(x) + \cos^2(y) = 1\) without using logarithms, we can proceed using implicit differentiation.

### Step 1: Differentiate the equation implicitly with respect to \(x\).
Given:
\[
\sin^2(x) + \cos^2(y) = 1
\]
Differentiate both sides with respect to \(x\):
\[
\frac{d}{dx} \left(\sin^2(x) + \cos^2(y)\right) = \frac{d}{dx} (1)
\]

### Step 2: Apply the chain rule.
For the first term \(\sin^2(x)\):
\[
\frac{d}{dx} \left(\sin^2(x)\right) = 2\sin(x) \cdot \cos(x)
\]
This simplifies to:
\[
2\sin(x)\cos(x) = \sin(2x)
\]

For the second term \(\cos^2(y)\):
\[
\frac{d}{dx} \left(\cos^2(y)\right) = 2\cos(y) \cdot (-\sin(y)) \cdot \frac{dy}{dx}
\]
This simplifies to:
\[
-2\sin(y)\cos(y) \cdot \frac{dy}{dx}
\]

### Step 3: Combine the derivatives.
The differentiated equation becomes:
\[
\sin(2x) - 2\sin(y)\cos(y) \cdot \frac{dy}{dx} = 0
\]

### Step 4: Solve for \(\frac{dy}{dx}\).
Rearrange the equation to isolate \(\frac{dy}{dx}\):
\[
-2\sin(y)\cos(y) \cdot \frac{dy}{dx} = -\sin(2x)
\]
Simplify:
\[
\frac{dy}{dx} = \frac{\sin(2x)}{2\sin(y)\cos(y)}
\]

This is the expression for \(\frac{dy}{dx}\).

### Do you want details or have any questions?

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### 5 Relative Questions:
1. How would the solution change if the equation involved a different trigonometric identity?
2. Can you find the second derivative \(\frac{d^2y}{dx^2}\) from this result?
3. What happens if \(x = 0\) or \(y = 0\) in this expression for \(\frac{dy}{dx}\)?
4. How would you solve this if the equation was \(\sin^2(x) + \sin^2(y) = 1\)?
5. How does implicit differentiation help in solving such equations without explicitly solving for \(y\)?

### Tip:
Always apply the chain rule carefully when dealing with composite functions during differentiation.

Discover how to find the derivative dy/dx of sin^2(x) + cos^2(y) = 1 using implicit differentiation without logarithms. This problem involves trigonometric identities and the chain rule in calculus. Suitable for advanced high school students and those studying calculus.

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