let ( f(x)=-2 e^{x} sin x )\n( f^{prime}(x)= )\nquestion help: video message instructor

Explanation:

Step1: Apply the product rule

The product rule states that if (y = uv), then (y^\prime=u^\prime v + uv^\prime). Let (u=-2e^{x}) and (v = \sin x). First, find (u^\prime) and (v^\prime). The derivative of (u=-2e^{x}) is (u^\prime=-2e^{x}) (since (\frac{d}{dx}(e^{x})=e^{x})), and the derivative of (v = \sin x) is (v^\prime=\cos x).

Step2: Substitute into the product rule formula

[ \begin{align*} f^\prime(x)&=u^\prime v+uv^\prime\ &=- 2e^{x}\cdot\sin x+(-2e^{x})\cdot\cos x\ &=-2e^{x}(\sin x+\cos x) \end{align*} ]

Answer:

(-2e^{x}(\sin x+\cos x))