Question

plzanswer.thhis question​

Answer 1

Step-by-step explanation:

I am considering ¢ to be alpha.

$y = 2 \sin(wx + \alpha ) \\ \frac{dy}{dx} = 2 \frac{d}{dx} ( \sin(wx + \alpha ) ) \\ = 2 \cos(wx + \alpha ) \times \frac{d}{dx} (wx + \alpha ) \\ = 2 \cos(wx + \alpha ) \times (w + 0) \\ = 2w \cos(wx + \alpha )$

Hence option 3 is the right answer.

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