y = 1/2arccosx, (-sqrt(2)/2, (3pi)/8) Find an equation of the tangent line to the graph of the function at the given point

Equation of a tangent line to the graph of function f  at point (x_0,y_0) is given by y=y_0+f'(x_0)(x-x_0).
The first step to finding equation of tangent line is to calculate the derivative of the given function.
y'=1/2cdot(-1/sqrt(1-x^2))=-1/(2sqrt(1-x^2))
Now we calculate the value of the derivative at the given point.
y'(-sqrt2/2)=-1/(2sqrt(1-(-sqrt2/2)^2))=-1/(2sqrt(1-1/2))=-1/(2sqrt2/2)=-1/sqrt2=-sqrt2/2
We now have everything needed to write the equation of the tangent line.
y=(3pi)/8-sqrt2/2(x+sqrt2/2)
y=-sqrt2/2x+(3pi-4)/8
Graph of the function along with the tangent line can be seen in the image below.                                                                     
https://en.wikipedia.org/wiki/Tangent

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