How do I find the derivative of the problems? Please explain?

Question

K(z) = 1-z/2z,
K'(-1),
K'(1),
K'(squre root of 2)

Please explain how they are done. I'm new to derivatives.

Answer 1

Rather than do anything complex, just split out the fraction (because (a+b)/c = a/c + b/c):
K(z) = (1-z)/2z
K(z) = 1/2z - z/2z = 1/(2z) - 1/2

Use the power rule: ax^b = (ab)x^(b-1)
d/dx 1/(2z) = d/dx 1/2 (z^-1) = 1/2 (-1)(z^-2) = -1/(2z^2)
d/dx 1/2 = 0 (derivative of a constant is always 0)

So, K'(z) = -1/(2z^2)
K'(-1) = -1/(2(-1)^2) = -1/(2) = -1/2
K'(1) = -1/(2(1)^2) = -1/2
K'(√2) = -1/(2(√2)^2) = -1/4

Answer 2

I'd set it up as k(z) = (1-z) (2z)^(-1) and use the product rule.

I get k' = (-3z + 1) / -(2z)^2