Question

Find the derivative of P(t)=(2)(e^(-2e)^(2t))).
Can someone please show me step by step how to solve this?

Answer 1

\[\LARGE P(t)=2e^{-2(e^{2t})}\]
Is that it or is it \[\LARGE P(t)=2e^{(-2e)^{2t}}\]

I just want to be sure I've got it right.

Answer 2

We need to know. @lana_lei

Answer 3

It's the second one.

Answer 4

Alright, try your best to do it. You're going to need to use the chain rule here a few times. Are you sure the negative sign is inside the parenthesis and not just the 2 like this:
\[\LARGE P(t)=2e^{-(2e)^{2t}}\]

Answer 5

There aren't any parenthesis in the question, but I just used them to separate each of the exponents. \[P(t) = 2\times e ^{-2e}^{2t}\]I'm really not sure where to start. Would I start by finding the compositions: g(x)=e^e and f(x) = -2e? But what about the 2t exponent?

Answer 6

If there aren't any parentheses, then the equation would be\[\Large P(t)=2e^{-2(e^{2t})}.\]The best way to differentiate this, would be to simply slowly apply the chain rule a few times. So the first equation you have, is\[f(u(t))=2e^{u(t)}\implies f'(u)=2e^{u(t)}\cdot u'(t)\]Now, replace \(u(t)\) with\[-2e^{2t},\]and \(u'(t)\) with\[-4e^{2t}.\]Simplify, and you should be done.

Answer 7

Thank you so much, this was taking me forever to do! Your explanation was really easy to understand!

Answer 8

You're welcome.