Differential of `sin^(2)(x^(2))" w.r.t. "x^(2)" is "sin2x^(2)`.

Answer

To find the differential of \( \sin^2(x^2) \) with respect to \( x^2 \), we will follow these steps:

### Step 1: Define the function
Let \( u = \sin^2(x^2) \).

### Step 2: Differentiate using the chain rule
To differentiate \( u \) with respect to \( x^2 \), we can use the chain rule. First, we will differentiate \( u \) with respect to \( x \) and then differentiate \( x^2 \) with respect to \( x \).

### Step 3: Differentiate \( u \) with respect to \( x \)
Using the chain rule:
\[ \frac{du}{dx} = 2\sin(x^2) \cdot \cos(x^2) \cdot \frac{d(x^2)}{dx} \]
Here, \( \frac{d(x^2)}{dx} = 2x \).

So,
\[ \frac{du}{dx} = 2\sin(x^2) \cdot \cos(x^2) \cdot 2x = 4x \sin(x^2) \cos(x^2) \]

### Step 4: Differentiate \( x^2 \) with respect to \( x \)
We know that:
\[ \frac{dv}{dx} = 2x \]
where \( v = x^2 \).

### Step 5: Find \( \frac{du}{dv} \)
Now, we need to find \( \frac{du}{dv} \):
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{4x \sin(x^2) \cos(x^2)}{2x} \]
The \( x \) terms cancel out (assuming \( x \neq 0 \)):
\[ \frac{du}{dv} = 2 \sin(x^2) \cos(x^2) \]

### Step 6: Use the double angle identity
We know that:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
Thus,
\[ \frac{du}{dv} = \sin(2x^2) \]

### Final Answer
Therefore, the differential of \( \sin^2(x^2) \) with respect to \( x^2 \) is:
\[ \frac{du}{dv} = \sin(2x^2) \]

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