Inactive Tutor answered 07/02/18
Tutor
New to Wyzant
Well, you can use the SECANT line to approximate the derivatives.
Basically, you calculate the slope of the secant line between each pair of
consecutive points.
For x=0,1,2,3,4,5,6,7,8,9,10,11 representing the month number, we
are interested in approximating f'(0.5), f'(1.5), f'(2.5) etc
Since x increases by 1 along the interval, the slope of the secant
line between each pair ( 0,382.44) , (1,383.69), (2,384.24) ... etc... (11,383.69)
is just the differences between each y-value.
To estimate the derivative, we can just cut the difference in half and add it
to the previous y-value.
Specifically, f'(x+1/2) is approximated by f(x) + 1/2( f(x+1) - f(x)) =
f(x) + 1/2 f(x+1) - 1/2f(x) =
1/2 f(x+1) + 1/2 f(x) =
[f(x+1) + f(x) ]/2
which is the AVERAGE of each y-value
The results are shown in the following table
x f(x) f(x+1)-f(x) approximation = [f(x)+f(x+1)]/2
0 382.44
1 383.69 1.25 383.065
2 384.24 0.55 383.965
3 386.25 2.01 385.245
4 386.39 0.14 386.32
5 385.86 -0.53 386.125
6 384.44 -1.42 385.15
7 381.77 -2.67 383.105
8 380.74 -1.03 381.255
9 380.8 0.06 380.77
10 382.33 1.53 381.565
11 383.69 1.36 383.01
1 383.69 1.25 383.065
2 384.24 0.55 383.965
3 386.25 2.01 385.245
4 386.39 0.14 386.32
5 385.86 -0.53 386.125
6 384.44 -1.42 385.15
7 381.77 -2.67 383.105
8 380.74 -1.03 381.255
9 380.8 0.06 380.77
10 382.33 1.53 381.565
11 383.69 1.36 383.01