Question

the equation of the tangent to y=cos^-1x at x=1/2

Answer 1

\[y^{'}=\frac{ 1 }{ -\sin(\cos^{-1}(x)) }\]Thus\[y'(\frac{ 1 }{ 2 })=\frac{ -1 }{ \sin(\cos^{-1}(\frac{ 1 }{ 2 })) }=\frac{ -1 }{ \sin(\frac{ \pi }{ 3 }) }=\frac{ -2 }{ \sqrt {3}}\]Then the equation of the line is \[\frac{ y-\frac{ \pi }{ 3 } }{ x-\frac{ 1 }{ 2} }=\frac{ -2 }{ \sqrt {3}}\]

Answer 2

thats one of the answers given but theres still more