How do you differentiate #f(x) = e ^ (2(e^(t) - 1)#?

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Answer 1 · 2016-07-15T11:10:24

typo?

#f(color{blue}{x}) = e ^ (2(e^(color{red}{t}) - 1)#

gotta typo in there? or trick question?

the trick with exponents is: #d/(du) ( e^(f(u)) )= (df)/(du) e^(f(u))#

Answer 2 · 2016-07-28T13:26:32

#f'(t) = 2e^(2(e^t-1)+t)#

We will assume that the function was meant to be #f(t)# rather than #f(x)#

Thus: #f(t) = e^(2(e^t-1))#

#f'(t) = e^(2(e^t-1)) * d/dt (2(e^t-1))#
(Standard Exponential and Chain rule)

#f'(t) = e^(2(e^t-1)) * 2(e^t - 0)#

#f'(t) = 2e^(2(e^t-1)+t)#