Daniel B. answered 11/18/20
A retired computer professional to teach math, physics
First, 200π/25 = 8π, and therefore you can remove the absolute value signs in case c) and d), which makes them identical to case a).
For case a)
f'(x) = 200cos(200x), and therefore f'(π/25) = 200 for cases a), c), d)
For case b), f(x) reverses polarity at point x = π/25.
By that I mean that when close to π/25, for x <π/25 sin(200x) is negative, and for x >π/25 sin(200x) is positive.
Therefore |sin(200x)| would be differentiable at x = π/25, only if the derivative of sin(200x) there were 0.
But since the derivative there is 200, |sin(200x)| is not differentiable at x = π/25.
It looks like the calculator was trying to give you 0.
Ty B.
I was instructed to use a calculator to find the answers. So are my answers correct for using a calculator?11/18/20