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What is the second derivative of #f(t) = (3e^-2t) - (5e^-t) #?

Answer 1

Christopher Agresta

#(d^2y)/(dt^2) = - 5/ e^t#

#f(t) = (3 e^(-2)t) - (5 e^(-t))# or

#f(t) = (3/ e^2 t) - (5 e^(-t))# or

#f^'(t) = (3/ e^2 ) - (5 e^(-t)*(-1))# or

#f^'(t) = (3/ e^2 ) + (5 e^(-t))# or

#(d^2y)/(dt^2) = 0 + (5 e^(-t)*(-1))# or

#(d^2y)/(dt^2) = - (5 e^(-t))# or

#(d^2y)/(dt^2) = - 5/ e^t# [Ans]

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Answer 2

Paisley Johnson

To find the second derivative of ( f(t) = 3e^{-2t} - 5e^{-t} ), you first need to find the first derivative and then differentiate it again.

First derivative of ( f(t) ): [ f'(t) = \frac{d}{dt}(3e^{-2t}) - \frac{d}{dt}(5e^{-t}) ]

[ f'(t) = -6e^{-2t} + 5e^{-t} ]

Now, differentiate ( f'(t) ) with respect to ( t ) to find the second derivative:

[ f''(t) = \frac{d}{dt}(-6e^{-2t} + 5e^{-t}) ]

[ f''(t) = \frac{d}{dt}(-6e^{-2t}) + \frac{d}{dt}(5e^{-t}) ]

[ f''(t) = 12e^{-2t} - 5e^{-t} ]

So, the second derivative of ( f(t) = 3e^{-2t} - 5e^{-t} ) is ( f''(t) = 12e^{-2t} - 5e^{-t} ).

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Answer from HIX Tutor

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