b) $\\frac{d}{dx}(2^{x}\\sin x)=$

Explanation:

Step1: Apply the product rule

The product rule states that if (y = u\cdot v), then (y^\prime=u^\prime v + uv^\prime). Let (u = 2^{x}) and (v=\sin x).

Step2: Differentiate (u = 2^{x})

The derivative of (a^{x}) with respect to (x) is (a^{x}\ln a). So, (u^\prime=\frac{d}{dx}(2^{x})=2^{x}\ln 2).

Step3: Differentiate (v=\sin x)

The derivative of (\sin x) with respect to (x) is (\cos x). So, (v^\prime=\frac{d}{dx}(\sin x)=\cos x).

Step4: Substitute (u^\prime), (u), (v^\prime), and (v) into the product rule formula

(\frac{d}{dx}(2^{x}\sin x)=2^{x}\ln 2\cdot\sin x+2^{x}\cdot\cos x=2^{x}(\ln 2\sin x+\cos x))

Answer:

(2^{x}(\ln 2\sin x+\cos x))