The derivative of the function 
F(theta)=(sin theta/1+cos theta)^2 

To find the derivative of the function \( F(\theta) = \left( \frac{\sin \theta}{1 + \cos \theta} \right)^2 \), we'll apply the chain rule and the quotient rule.

### Step-by-step process:

1. **Rewrite the function:**

   \[
   F(\theta) = \left( \frac{\sin \theta}{1 + \cos \theta} \right)^2
   \]

2. **Use the chain rule:**

   The derivative of \( \left( g(\theta) \right)^2 \) is \( 2g(\theta) \cdot g'(\theta) \). In our case, \( g(\theta) = \frac{\sin \theta}{1 + \cos \theta} \). 

   So, we need to find the derivative of \( g(\theta) = \frac{\sin \theta}{1 + \cos \theta} \) using the quotient rule.

3. **Apply the quotient rule:**

   The quotient rule states:
   
   \[
   \frac{d}{d\theta} \left( \frac{f(\theta)}{h(\theta)} \right) = \frac{f'(\theta) h(\theta) - f(\theta) h'(\theta)}{h(\theta)^2}
   \]
   
   Here, \( f(\theta) = \sin \theta \) and \( h(\theta) = 1 + \cos \theta \).
   
   - \( f'(\theta) = \cos \theta \)
   - \( h'(\theta) = -\sin \theta \)
   
   Therefore, the derivative of \( g(\theta) = \frac{\sin \theta}{1 + \cos \theta} \) is:

   \[
   g'(\theta) = \frac{ \cos \theta (1 + \cos \theta) - \sin \theta (-\sin \theta) }{(1 + \cos \theta)^2}
   \]
   
   Simplifying:

   \[
   g'(\theta) = \frac{\cos \theta (1 + \cos \theta) + \sin^2 \theta}{(1 + \cos \theta)^2}
   \]

   Further simplifying:

   \[
   g'(\theta) = \frac{\cos \theta + \cos^2 \theta + \sin^2 \theta}{(1 + \cos \theta)^2}
   \]

   Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we get:

   \[
   g'(\theta) = \frac{\cos \theta + 1}{(1 + \cos \theta)^2}
   \]

4. **Multiply by the chain rule factor:**

   Now apply the chain rule:
   
   \[
   F'(\theta) = 2 \left( \frac{\sin \theta}{1 + \cos \theta} \right) \cdot \frac{\cos \theta + 1}{(1 + \cos \theta)^2}
   \]

5. **Final expression:**

   Simplifying the above:

   \[
   F'(\theta) = \frac{2 \sin \theta (\cos \theta + 1)}{(1 + \cos \theta)^3}
   \]

Thus, the derivative of \( F(\theta) \) is:

\[
F'(\theta) = \frac{2 \sin \theta (1 + \cos \theta)}{(1 + \cos \theta)^3}
\]

Or further simplified:

\[
F'(\theta) = \frac{2 \sin \theta}{(1 + \cos \theta)^2}
\]

Would you like a detailed breakdown of any of the steps?

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Here are five related questions for you:

1. What is the second derivative of the function \( F(\theta) \)?
2. How would the derivative change if the numerator was \( \cos \theta \) instead of \( \sin \theta \)?
3. How does the chain rule apply in the case of functions raised to powers?
4. Can the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) simplify similar problems?
5. How can you find the derivative of a product of two trigonometric functions?

**Tip:** Always check for trigonometric identities that can simplify the expressions before applying calculus techniques.

The derivative of the function F(θ) = (sin(θ) / (1 + cos(θ)))^2.

Learn how to find the derivative of the function F(θ) = (sin(θ) / (1 + cos(θ)))^2 using the chain rule and quotient rule. This step-by-step solution covers important concepts in calculus, including the differentiation of trigonometric functions and the use of trigonometric identities.

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