# How do you differentiate this...?

##
#e^(tsin2t)#

##### 1 Answer

Jun 10, 2018

#### Explanation:

We want to find the derivative of

#y=e^(t*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)#