Find the derivative of `3sinx+2sinalpha`, where `'alpha'` is a constant.

Answer

To find the derivative of the function \( y = 3\sin x + 2\sin \alpha \), where \( \alpha \) is a constant, we can follow these steps:

### Step 1: Identify the function
We start with the function:
\[ y = 3\sin x + 2\sin \alpha \]

### Step 2: Differentiate the function
We differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(3\sin x) + \frac{d}{dx}(2\sin \alpha) \]

### Step 3: Apply the derivative rules
1. The derivative of \( 3\sin x \) is \( 3\cos x \) (using the derivative of sine).
2. The derivative of \( 2\sin \alpha \) is \( 0 \) because \( \alpha \) is a constant and the derivative of a constant is zero.

Putting it all together, we have:
\[ \frac{dy}{dx} = 3\cos x + 0 \]
Thus, we simplify to:
\[ \frac{dy}{dx} = 3\cos x \]

### Final Answer
The derivative of \( 3\sin x + 2\sin \alpha \) is:
\[ \frac{dy}{dx} = 3\cos x \]

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