# Calculus

How do you find the first and second derivative of the following function?

P(t)=20/(1+3e^-0.02t)

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1. P(t) = 20/(1+3e^-0.02t)
= 20(1+3e^-.02t)^-1

P'(t) = 20(3e^-0.02t)(-0.02) * (-1)(1+3e^-0.02t)^-2
= 1.2e^-.02t (1+3e^-.02t)^-2

P"(t) =
1.2e^-.02t(-.02)(1+3e^-.02t)^-2
1.2e^-.02t (-2)(1+3e^-.02t)^-3 (-.06e^-.02t)
which, after suitable rearranging is
= 0.024e^-.02t (3e^-.02t - 1)/(1+3e^-.02t)^3

better check my algebra

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2. P(t)=20(1+3e^-.02t)^-1

P'=-20(1+3e^-.02t)^-2 * (-.06e^-.02t)

P"= 40(1+3e^-.02t)^-3 * (-.06e^-.02t)*(.0012e^-.02t)

check all that.

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👨‍🏫
bobpursley
3. write it as

P(t) = 20 (1 + 3e^(-.02t)^-1

P ' (t) = -20(1+3e^-.02t)^-2 (-.02 e^(-.02t) )

now use the product rule to get the 2nd derivatives
I suggest cleaning up the first deriv. a bit

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4. go with Steve's
I forgot to multiply the .02 by the 3 in front

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