The second derivative of the first function is s"(x) = = -[8e^(x/5) - 2]e^(x/5)/25(e^(x/5) + 2)³.
The second derivative of the second function is; k''(t) = (36t² - 6)e^(-3t²)
How to find the second derivative?
1) We are given the function;
s(x) = 4/(1 + 2e^(-0.2x)
First derivative is;
ds/dx = [8e^(x/5)]/5(e^(x/5) + 2)²
Second Derivative is;
d²s/dx² = -[8e^(x/5) - 2]e^(x/5)/25(e^(x/5) + 2)³
At x = 0;
d²s/dx² = -[8e^(0/5) - 2]e^(0/5)/25(e^(0/5) + 2)³
d²s/dx² = 8/675
B) We are given the function;
k(t) = e^(-3t²)
The first derivative is;
dk/dt = -6te^(-3t²)
The second derivative is;
k''(t) = (36t² - 6)e^(-3t²)
At t = 1, we have;
k''(t) = (36(1)² - 6)e^(-3(1²))
k"(t) = 1.494
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