`Ify= (sin ^(2) x )/( 1+cos^(2) x) ,then (dy)/(dx) =`

Options

A. ` (-sin 2x )/( (1+ cos ^(2)x) ^(2))`
B. ` (sin 2x ) /((1+cos ^(2)x) ^(2) )`
C. ` (-2sin 2x )/( (1+cos ^(2)x))`
D. ` (2sin 2x)/( (1+cos ^(2) x) ^(2))`

Answer

To find the derivative of the function \( y = \frac{\sin^2 x}{1 + \cos^2 x} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( y = \frac{f(x)}{g(x)} \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]

### Step-by-step solution:

1. **Identify \( f(x) \) and \( g(x) \)**:
- Let \( f(x) = \sin^2 x \)
- Let \( g(x) = 1 + \cos^2 x \)

2. **Differentiate \( f(x) \)**:
- Using the chain rule, \( f'(x) = 2 \sin x \cdot \cos x \) (since the derivative of \( \sin^2 x \) is \( 2 \sin x \cdot \cos x \))

3. **Differentiate \( g(x) \)**:
- \( g'(x) = 0 + 2 \cos x \cdot (-\sin x) = -2 \cos x \sin x \) (the derivative of \( \cos^2 x \) is \( -2 \cos x \sin x \))

4. **Apply the quotient rule**:
- Now substitute \( f(x) \), \( g(x) \), \( f'(x) \), and \( g'(x) \) into the quotient rule formula:
\[ \frac{dy}{dx} = \frac{(2 \sin x \cos x)(1 + \cos^2 x) - (\sin^2 x)(-2 \cos x \sin x)}{(1 + \cos^2 x)^2} \]

5. **Simplify the numerator**:
- The numerator becomes:
\[ 2 \sin x \cos x (1 + \cos^2 x) + 2 \sin^3 x \cos x \]
- Factor out \( 2 \sin x \cos x \):
\[ = 2 \sin x \cos x \left(1 + \cos^2 x + \sin^2 x\right) \]

6. **Use the Pythagorean identity**:
- Recall that \( \sin^2 x + \cos^2 x = 1 \):
\[ 1 + \cos^2 x + \sin^2 x = 1 + 1 = 2 \]

7. **Final expression for \( \frac{dy}{dx} \)**:
- Substitute back into the expression:
\[ \frac{dy}{dx} = \frac{2 \sin x \cos x \cdot 2}{(1 + \cos^2 x)^2} = \frac{4 \sin x \cos x}{(1 + \cos^2 x)^2} \]

8. **Use the double angle identity**:
- Recall that \( 2 \sin x \cos x = \sin(2x) \):
\[ \frac{dy}{dx} = \frac{2 \sin(2x)}{(1 + \cos^2 x)^2} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{2 \sin(2x)}{(1 + \cos^2 x)^2} \]