How do you differentiate this...?

Question

#e^(tsin2t)#

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Answer 1 · 2018-06-10T07:08:04

#dy/dt=(sin(2t)+2tcos(2t))e^(t*sin(2t)#

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)=t*sin(2t)#, and by the product rule

#color(red)(f'(t)=sin(2t)+2tcos(2t)#

Thus

#dy/dt=(sin(2t)+2tcos(2t))e^(t*sin(2t)#