If `f(x)=sin^(2)x^(2)`, then what is `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))`

Answer

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

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

### Step 2: Differentiate the outer function
Using the power rule, the derivative of \( u^2 \) with respect to \( u \) is:
\[ \frac{d}{du}(u^2) = 2u \]
Substituting back \( u = \sin(x^2) \), we have:
\[ \frac{d}{du}(u^2) = 2\sin(x^2) \]

### Step 3: Differentiate the inner function
Next, we differentiate the inner function \( u = \sin(v) \) where \( v = x^2 \):
\[ \frac{d}{dv}(\sin(v)) = \cos(v) \]
Substituting back \( v = x^2 \), we have:
\[ \frac{d}{dv}(\sin(x^2)) = \cos(x^2) \]

### Step 4: Differentiate the innermost function
Now, we differentiate \( v = x^2 \):
\[ \frac{d}{dx}(x^2) = 2x \]

### Step 5: Apply the chain rule
Now we apply the chain rule:
\[ f'(x) = \frac{d}{du}(u^2) \cdot \frac{d}{dv}(\sin(v)) \cdot \frac{d}{dx}(x^2) \]
Substituting the derivatives we calculated:
\[ f'(x) = 2\sin(x^2) \cdot \cos(x^2) \cdot 2x \]
Simplifying this gives:
\[ f'(x) = 4x \sin(x^2) \cos(x^2) \]

### Final Answer
Thus, the derivative \( f'(x) \) is:
\[ f'(x) = 4x \sin(x^2) \cos(x^2) \]

---