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

1 Answer

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..