A cylindrical pipe has a circumference of 16cm
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9514 1404 393
Answer:
see attached
Step-by-step explanation:
One way to approximate the derivative at a point is by finding the slope of the secant line between points on either side. That is what is done in the attached spreadsheet.
f'(0.1) ≈ (f(0.2) -f(0.0))/(0.2 -0.0) = -5 . . . for example
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Another way to approximate the derivative is to write a polynomial function that goes through the points (all, or some subset around the point of interest), and use the derivative of that polynomial function.
These points are reasonably approximated by a cubic polynomial. The derivative of that polynomial at the points of interest is given in the table in the second attachment. (f1 is a rounding of the derivative function f')
