If `f(x)=sin^(2)x^(2),` then what f'(x) equal to?

Options

A. `4xsin(x^(2))cos(x^(2))`
B. `2sin(x^(2))cos(x^(2))`
C. `4sin(x^(2))sin^(2)x`
D. `2xcos^(2)(x^(2))`

Explanation

To find the derivative of the function \( f(x) = \sin^2(x^2) \), we will apply the chain rule. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
We can identify the outer function and the inner function:
- Outer function: \( u = \sin^2(v) \) where \( v = x^2 \)
- Inner function: \( v = x^2 \)

### Step 2: Differentiate the outer function
Using the chain rule, the derivative of \( u = \sin^2(v) \) is:
\[ \frac{du}{dv} = 2\sin(v)\cos(v) = \sin(2v) \]
This is derived from the identity \( \sin^2(v) = \frac{1 - \cos(2v)}{2} \), but we can also use the product rule directly.

### Step 3: Differentiate the inner function
Now, we differentiate the inner function \( v = x^2 \):
\[ \frac{dv}{dx} = 2x \]

### Step 4: Apply the chain rule
Now, we can apply the chain rule:
\[ \frac{df}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} \]
Substituting the derivatives we found:
\[ \frac{df}{dx} = \sin(2v) \cdot 2x \]
Substituting back \( v = x^2 \):
\[ \frac{df}{dx} = \sin(2x^2) \cdot 2x \]

### Step 5: Final expression
Thus, the derivative \( f'(x) \) is:
\[ f'(x) = 2x \sin(2x^2) \]

### Summary
The final answer is:
\[ f'(x) = 2x \sin(2x^2) \]