Answer
$F'(t)=e^{tsin2t}.[ 2t cos(2t)+sin 2t]$
Work Step by Step
Need to solve $F'(t)=\frac{d}{dt}(e^{tsin2t)}$
Apply chain rule.
Suppose $u = tsin2t$
Then, $F'(t)=\frac{d}{dt}(e^{u})$
Also we can write as follows:
$F'(t)=\frac{dF}{du}\times\frac{du}{dF}$
First determine $\frac{dF}{du}$
$F'(t)=\frac{dF}{dt}=\frac{dF}{du}\times\frac{du}{dt}$ ...(1)
Thus, $\frac{dF}{du}=e^{u}$
and $\frac{du}{dt} =2t cos2t+sin2t$
From equation (1) , we have $F'(t)=\frac{dF}{dt}=\frac{dF}
{du}\times\frac{du}{dt}$
$F'(t)=e^{u}.[ 2t cos(2t)+sin 2t]$
Hence, $F'(t)=e^{tsin2t}.[ 2t cos(2t)+sin 2t]$
