Question

Differentiate Phi(t) = exp{ λ(e^t − 1)},

Answer 1

with respect to t?

Answer 2

you there?

Answer 3

does exp mean e^()

Answer 4

yeps!

Answer 5

@tomo

Answer 6

exp{ λ(e^t − 1)},

y = lambda. just easier to type

= e^(y*e^t - y)

break up the e

= [e^(y*e^t)]*(e^-y)

Answer 7

okkk

Answer 8

you can type it into wolfram and see what it spits out. maybe that will help you work backward and see how they arrive at the answer.

Answer 9

http://www.wolframalpha.com/input/?i=d%2Fdt%5Be%5E%28y*e%5Et%5D

Answer 10

\[\large y=e^{y(e^t - 1)}\]
\[\large \frac{dy}{dt}=e^{y(e^t - 1)}[(e^t-1)(\frac{dy}{dt})+y(e^t)]\]
\[\large \frac{dy}{dt}=e^{y(e^t - 1)}(e^t-1)(\frac{dy}{dt})+e^{y(e^t - 1)}ye^t\]
\[\large \frac{dy}{dt}-e^{y(e^t - 1)}\frac{dy}{dt}=e^{y(e^t - 1)}ye^t\]
\[\large \frac{dy}{dt}[1-e^{y(e^t - 1)}]=e^{y(e^t - 1)}ye^t\]
\[\large \frac{dy}{dt}=\frac{e^{y(e^t - 1)}ye^t}{1-e^{y(e^t - 1)}}\]
\[\large \frac{dy}{dt}=\frac{ye^{y(e^t-1)+t}}{1-e^{y(e^t - 1)}}\]