find the derivative of y with respect to t.\ny = e^{(3\\sin t + \\ln t)}\n\\frac{dy}{dt} = \\square

Explanation:

Step1: Apply the chain rule

Let (u = 3\sin t+\ln t), then (y = e^{u}). The chain rule states that (\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}). First, (\frac{dy}{du}=e^{u}) (since the derivative of (e^{x}) with respect to (x) is (e^{x})).

Step2: Differentiate (u) with respect to (t)

Differentiate (u = 3\sin t+\ln t) term - by - term. The derivative of (3\sin t) with respect to (t) is (3\cos t) (using the formula (\frac{d}{dt}(\sin t)=\cos t)), and the derivative of (\ln t) with respect to (t) is (\frac{1}{t}) (using the formula (\frac{d}{dt}(\ln t)=\frac{1}{t})). So, (\frac{du}{dt}=3\cos t+\frac{1}{t}).

Step3: Combine the results

Since (\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}) and (y = e^{u}), (u = 3\sin t+\ln t), (\frac{dy}{dt}=e^{(3\sin t+\ln t)}\left(3\cos t+\frac{1}{t}\right)).

Answer:

(e^{(3\sin t+\ln t)}\left(3\cos t+\frac{1}{t}\right))