Prove that `f(x)=sinx+sqrt(3)cosx` has maximum value at `x=pi/6dot`

Explanation

To prove that the function \( f(x) = \sin x + \sqrt{3} \cos x \) has a maximum value at \( x = \frac{\pi}{6} \), we will follow these steps:

### Step 1: Differentiate the function
We first find the derivative of the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin x + \sqrt{3} \cos x) = \cos x - \sqrt{3} \sin x \]

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