Thread: Differentiating E

Originally Posted by Xavier20

y' = [sin(2t)tcos(2t)*2]e^(tsin(2t))

Yeah. I forgot to add the + sign between the sin(2t) and tcos(2t) and didn't bring the *2 out infront of the tcos(2t) in my chain rule.

Thanks for sorting that out though.

It should be:

y' = [sin(2t) + 2tcos(2t)]e^(tsin(2t))

correct. i guess you're a genius with derivatives involving e now, but practice a few more problems to make sure

good luck!

Thanks a lot. =)