# Find Derivative of f(x) = arccos(cos(x)) and graph it

A calculus tutorial on how to find the first derivative of f(x) = arccos(cos(x)) and graph f and f' for x in R.

 Since the domain of f is R and cos(x) is periodic, then f(x) = arccos(cos(x)) is also a periodic function. As x increases from 0 to Pi, cos(x) decreases from 1 to -1 and arccos(cos(x)) increases from 0 to Pi. In fact for x in [0 , pi] arccos(cos(x)) = x. As x increases from [Pi , 2pi], cos(x) increases from -1 to 1 and arccos(cos(x)) decreases from Pi to 0. Since cos(x) has a period of 2Pi, arccos(cos(x)) also has a period of 2Pi. The graph below shows the graphs of arccos(cos(x)) and sin(x) from 0 to 2Pi. The graph below shows the graphs of arccos(cos(x)) and cos(x) over 3 periods. Domain of f: (-infinity , + infinity) Range of f: [0 , pi] Derivative of f(x) = arccos(cos(x)) f(x) is a composite function and the derivative is computed using the chain rule as follows: Let u = cos(x) Hence f(x) = arccos(u(x)) Apply the chain rule of differentiation f'(x) = du/dx d(arccos(u))/du = -sin(x) * (- 1 / sqrt(1 - u2)) = sin(x) * 1 / (1 - cos2(x)) = sin(x) / sqrt(sin2(x)) = sin(x) / | sin(x) | Below is shown arccos(cos(x)) in red and its derivative in blue. Note that the derivative is undefined for values of x for which sin(x) = 0, which means at x = k*pi, where k is an integer. For these same values of x, arccos(cos(x)) has either a maximum value equal to pi or a minimum value equal to 0. Note that although arccos(cos(x)) is continuous for all values of x, its derivative is undefined at x = k*pi. More on differentiation and derivatives