Question

How do you find an equation of the tangent line to the curve at the given point `y=3arccos(x/2)` and `(1,pi)`?

Answer 1

`y=-sqrt3*x+sqrt3+pi`

Explanation:

From the given curve `y=3*arccos(x/2)` and point `(1, pi)`

Find the first derivative by the formula

`d/dx(arccos u)=(-1)/sqrt(1-u^2)*d/dx(u)`

`(dy)/dx=d/dx(3*arccos(x/2))=3*(-1)/sqrt(1-(x/2)^2)*d/dx(x/2)`

`(dy)/dx=-3/(1/2*sqrt(4-x^2))*(1/2)`

at `x=1`

`(dy)/dx=-3/(1/2*sqrt(4-1^2))*(1/2)`

`(dy)/dx=-3/sqrt3=-sqrt3""""" "the" "slope"`

The Tangent line:

By using point-slope formula with slope `m=-sqrt3` and given point `(x_1, y_1)=(1, pi)`

`y-y_1=m*(x-x_1)`

`y-pi=-sqrt3*(x-1)`

`y=-sqrt3*x+sqrt3+pi`

I hope the explanation is useful...God bless..