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
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Thanks a lot. =)
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