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.




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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.

Graph of cos(x) and arccos(cos(x) over one period)


The graph below shows the graphs of arccos(cos(x)) and cos(x) over 3 periods.

Graph of cos(x) and arccos(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.

Graph of arccos(cos(x)) and its first derivative




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