Find an equation of the tangent line to the curve $y = 8 \arccos \left( \frac { x } { 2 } \right)$ at the point $( 1 , \pi )$.
A) $y = \frac { \sqrt { 8 } } { 3 } x + 6 \pi - \frac { \sqrt { 8 } } { 3 }$
B)$y = - \frac { \sqrt { 3 } } { 8 } x + \pi + 6 \frac { \sqrt { 3 } } { 8 }$
C)$y = - \frac { 8 } { \sqrt { 3 } } x + \pi + \frac { 8 } { \sqrt { 3 } }$
D)$y = \frac { 8 } { \sqrt { 3 } } x + 3 \pi - \frac { 8 } { \sqrt { 3 } }$
E)$y = \frac { \sqrt { 3 } } { 8 } x + \pi - 6 \frac { \sqrt { 3 } } { 8 }$

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The Answer of Find an equation of the tangent line to the curve $y = 8 \arccos \left( \frac { x } { 2 } \right)$ at the point $( 1 , \pi )$.
A) $y = \frac { \sqrt { 8 } } { 3 } x + 6 \pi - \frac { \sqrt { 8 } } { 3 }$
B)$y = - \frac { \sqrt { 3 } } { 8 } x + \pi + 6 \frac { \sqrt { 3 } } { 8 }$
C)$y = - \frac { 8 } { \sqrt { 3 } } x + \pi + \frac { 8 } { \sqrt { 3 } }$
D)$y = \frac { 8 } { \sqrt { 3 } } x + 3 \pi - \frac { 8 } { \sqrt { 3 } }$
E)$y = \frac { \sqrt { 3 } } { 8 } x + \pi - 6 \frac { \sqrt { 3 } } { 8 }$

Find an Equation of the Tangent Line to the Curve y=8arccos⁡(x2)y = 8 \arccos \left( \frac { x } { 2 } \right)y=8arccos(2x​)

Find an Equation of the Tangent Line to the Curve $y = 8 \arccos \left( \frac { x } { 2 } \right)$