Find the Derivative P(t) = 2*e^(-2e)^(2t) P prime (t) =

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Find the Derivative P(t) = 2*e^(-2e)^(2t) P prime (t) =

Mathematics
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By using the chain rule we get\[P'(t)=2e^{(-2e)^{2t}}*-2e^{2t}*2\] It's a tedious one, but if you take it slow and break it down and it's not too bad. Chain rules within chain rules are always annoying
yeah, it marked it as wrong ^^^^^
Interesting.

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Other answers:

If the exponent is `really` (-2t)^(2t), then this one is going to be be a lil more frustrating since the -2 is included lol
yeah ^
would we make a substitution? \[u=-2e ^{2t}\]
See that's different than what you initially wrote :P
\[\large\rm y=a^x\]Recall that for differentiating exponentials:\[\large\rm y'=a^x(\ln a)\]So then,\[\large\rm \frac{d}{dt}(-2e)^{2t}=(-2e)^{2t}\cdot \ln(-2e)\cdot (2)\]Oh but then log of -2e isn't defined... Yah having a negative base is uhh... no bueno :U hmm
yeaah, see i tried wolfphram alpha and it gave me an imaginary number???
Do you have a picture of the problem or something?
Ugh :( Ya that's not what you wrote out on top -_-
sorry i tried my best
Then Chris's derivative is correct :) We just gotta be careful with the brackets.
P(t)=2*e^(-2e^(2t)) P'(t)=2e(-2e^(2t))*(-2)e^(2t)*2 Try that maybe? :d Do you only get a limited number of "guesses"?
I'm on my last one, but oh well only got 3 mins left
oh XD
oops i made a typo :O hopefully you caught that. umm the first part of it
P'(t)=2e^(-2e^(2t))*(-2)e^(2t)*2 there we go, forgot the carot on the first exponential >.<
carot hahaaha
caret* :3 i dunno, whatever this thing is called "^" hehe
oh haha it sorta looks like a carrot anyway
ouch lol was a second late, when i hit submit it turned to 12 am :(
automatically took me home
aw :P does it show solution?
lame t.t
it doesn't :(
im gonna get on it on office hours, thank guys!

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