If $$s = \sqrt {{t^2} + 1} $$, then $${{{d^2}s} \over {d{t^2}}}$$ is equal to

Let $$f(x) = {(\left| x \right| - \left| {x - 1} \right|)^2}$$

What is f'(x) equal to when x > 1 ?

Let $$f(x) = {(\left| x \right| - \left| {x - 1} \right|)^2}$$

What is f'(x) equal to when 0 < x < 1 ?

Let $$f(x) = {(\left| x \right| - \left| {x - 1} \right|)^2}$$

Which of the following equations is/are correct?

I. f($$-$$2) = f(5)

II. f''($$-$$2) + f''(0.5) + f''(3) = 4

Select the correct answer using the code given below.