`y = 4arccos(x-1) , (1, 2pi)` Find an equation of the tangent line to the graph of the function at the given point

The equation of the tangent line has the form

`(y-y_0) = (x-x_0)*y'(x_0).`

Here `x_0 = 1` and `y_0 = 2 pi.`  This point is really on the graph, because `y_0 = 2pi` is really equal to  `4arccos(x_0 - 1) = 4arccos(0) = 4*pi/2 = 2pi.`

The derivative is  `y'(x) = -4/sqrt(1 - (x - 1)^2).` At `x = 1` it is equal to `-4,` and the equation of the tangent line is

`y - 2pi = (x - 1)*(-4), or y = -4x + 4 + 2pi.`