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

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

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

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.

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

go with Steve's

I forgot to multiply the .02 by the 3 in front

To find the first and second derivatives of the given function P(t), we can follow these steps:

Step 1: Rewrite the function in a more convenient form:
P(t) = 20/(1 + 3e^(-0.02t))

Step 2: Differentiate the function with respect to t using the quotient rule:
The quotient rule states that if we have a function of the form f(t) = g(t)/h(t), where g(t) and h(t) are both functions of t, then the derivative of f(t) is given by:

f'(t) = (g'(t) * h(t) - g(t) * h'(t)) / [h(t)]^2

In our case, g(t) = 20 and h(t) = (1 + 3e^(-0.02t)).
Now, let's differentiate the function P(t):

P'(t) = [20 * (1 + 3e^(-0.02t))' - 20' * (1 + 3e^(-0.02t))] / [(1 + 3e^(-0.02t))^2]

Step 3: Simplify the derivatives:

The derivative of (1 + 3e^(-0.02t)) can be found using the chain rule:
d/dt[e^u] = (du/dt) * e^u, where u = -0.02t

So, (1 + 3e^(-0.02t))' = 3 * (-0.02) * e^(-0.02t) = -0.06e^(-0.02t)

The derivative of a constant (20 in our case) is 0. Therefore, 20' = 0.

Substituting these values back into the derivative equation, we get:

P'(t) = [20 * (-0.06e^(-0.02t)) - 0 * (1 + 3e^(-0.02t))] / [(1 + 3e^(-0.02t))^2]
= -1.2e^(-0.02t) / (1 + 3e^(-0.02t))^2

So, we have found the first derivative of P(t) with respect to t.

Step 4: Find the second derivative:

To find the second derivative, we need to differentiate P'(t) with respect to t once again.
Let's denote P'(t) as Q(t) for simplicity.

Q(t) = -1.2e^(-0.02t) / (1 + 3e^(-0.02t))^2

Using the quotient rule, we can find Q'(t):

Q'(t) = [(-1.2e^(-0.02t))' * (1 + 3e^(-0.02t))^2 - (-1.2e^(-0.02t)) * (1 + 3e^(-0.02t))^2'] / [(1 + 3e^(-0.02t))^2]^2

Now, differentiate both terms within Q'(t):

(-1.2e^(-0.02t))' = 1.2 * (-0.02) * e^(-0.02t) = -0.024e^(-0.02t)

(1 + 3e^(-0.02t))^2' = 2 * (1 + 3e^(-0.02t)) * (3 * (-0.02) * e^(-0.02t)) = -0.36e^(-0.02t) * (1 + 3e^(-0.02t))

Substituting these values back into Q'(t), we get:

Q'(t) = [-0.024e^(-0.02t) * (1 + 3e^(-0.02t))^2 - (-1.2e^(-0.02t)) * (-0.36e^(-0.02t) * (1 + 3e^(-0.02t)))] / [(1 + 3e^(-0.02t))^2]^2
= (-0.024e^(-0.02t) * (1 + 3e^(-0.02t))^2 + 0.432e^(-0.02t) * (1 + 3e^(-0.02t))^2) / [(1 + 3e^(-0.02t))^2]^2
= (0.408e^(-0.02t) * (1 + 3e^(-0.02t))^2) / [(1 + 3e^(-0.02t))^2]^2
= 0.408e^(-0.02t) / (1 + 3e^(-0.02t))^2

So, we have found the second derivative of P(t) with respect to t.

To summarize:
P'(t) = -1.2e^(-0.02t) / (1 + 3e^(-0.02t))^2
P''(t) = 0.408e^(-0.02t) / (1 + 3e^(-0.02t))^2