Differentiate `sin^(2) (theta^(2)+1) w.r.t. theta^(2)`.

Answer

To differentiate \( \sin^2(\theta^2 + 1) \) with respect to \( \theta^2 \), we will follow these steps:

### Step 1: Identify the function
Let \( H = \sin^2(\theta^2 + 1) \) and \( g = \theta^2 \).

### Step 2: Differentiate \( H \) with respect to \( g \)
Using the chain rule, we know that:
\[ \frac{dH}{dg} = \frac{dH}{d(\theta^2)} = \frac{d}{d(\theta^2)}(\sin^2(\theta^2 + 1)) \]

### Step 3: Apply the chain rule
Using the chain rule, we differentiate \( H \):
\[ \frac{dH}{dg} = 2\sin(\theta^2 + 1) \cdot \cos(\theta^2 + 1) \cdot \frac{d}{d(\theta^2)}(\theta^2 + 1) \]

### Step 4: Differentiate the inner function
The derivative of \( \theta^2 + 1 \) with respect to \( \theta^2 \) is:
\[ \frac{d}{d(\theta^2)}(\theta^2 + 1) = 1 \]

### Step 5: Combine the results
Now substituting back, we have:
\[ \frac{dH}{dg} = 2\sin(\theta^2 + 1) \cdot \cos(\theta^2 + 1) \cdot 1 \]
Thus,
\[ \frac{dH}{dg} = 2\sin(\theta^2 + 1) \cos(\theta^2 + 1) \]

### Step 6: Use the double angle identity
We can use the double angle identity for sine:
\[ \sin(2x) = 2\sin(x)\cos(x) \]
So we can write:
\[ \frac{dH}{dg} = \sin(2(\theta^2 + 1)) \]

### Final Answer
Thus, the derivative of \( \sin^2(\theta^2 + 1) \) with respect to \( \theta^2 \) is:
\[ \frac{d}{d(\theta^2)}(\sin^2(\theta^2 + 1)) = \sin(2(\theta^2 + 1)) \]

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