How do you differentiate this...?
e^(tsin2t)etsin2t
1 Answer
Jun 10, 2018
Explanation:
We want to find the derivative of
y=e^(t*sin(2t))y=et⋅sin(2t)
By the chain rule we have
color(blue)(y=e^(f(x))=>dy/dx=f'(x)e^(f(x))
So for your problem
color(red)(f'(t)=sin(2t)+2tcos(2t)
Thus
dy/dt=(sin(2t)+2tcos(2t))e^(t*sin(2t)