Question

find the derivative of the function.

$f(t)=e^{6tsin(2t)}$

Explanation:

Step1: Apply the chain rule

Let \(u = 6t\sin(2t)\). Then \(F(t)=e^{u}\). By the chain rule \(\frac{dF}{dt}=\frac{dF}{du}\cdot\frac{du}{dt}\). Since \(\frac{d}{du}(e^{u}) = e^{u}\), we first find \(\frac{du}{dt}\).

Step2: Apply the product rule to find \(\frac{du}{dt}\)

For \(u = 6t\sin(2t)\), by the product rule \((uv)^\prime=u^\prime v + uv^\prime\) where \(u = 6t\), \(u^\prime=6\) and \(v=\sin(2t)\), \(v^\prime = 2\cos(2t)\). So \(\frac{du}{dt}=6\sin(2t)+6t\cdot2\cos(2t)=6\sin(2t)+12t\cos(2t)\)

Step3: Combine results

Since \(\frac{dF}{dt}=e^{u}\cdot\frac{du}{dt}\) and \(u = 6t\sin(2t)\), \(\frac{dF}{dt}=e^{6t\sin(2t)}(6\sin(2t)+12t\cos(2t)) = 6e^{6t\sin(2t)}(\sin(2t)+2t\cos(2t))\)

Answer:

\(6e^{6t\sin(2t)}(\sin(2t)+2t\cos(2t))\)