From the given curve y=3*arccos(x/2)y=3⋅arccos(x2) and point (1, pi)(1,π)
Find the first derivative by the formula
d/dx(arccos u)=(-1)/sqrt(1-u^2)*d/dx(u)ddx(arccosu)=−1√1−u2⋅ddx(u)
(dy)/dx=d/dx(3*arccos(x/2))=3*(-1)/sqrt(1-(x/2)^2)*d/dx(x/2)dydx=ddx(3⋅arccos(x2))=3⋅−1√1−(x2)2⋅ddx(x2)
(dy)/dx=-3/(1/2*sqrt(4-x^2))*(1/2)dydx=−312⋅√4−x2⋅(12)
at x=1x=1
(dy)/dx=-3/(1/2*sqrt(4-1^2))*(1/2)dydx=−312⋅√4−12⋅(12)
(dy)/dx=-3/sqrt3=-sqrt3""""" "the" "slope"dydx=−3√3=−√3 the slope
The Tangent line:
By using point-slope formula with slope m=-sqrt3m=−√3 and given point (x_1, y_1)=(1, pi)(x1,y1)=(1,π)
y-y_1=m*(x-x_1)y−y1=m⋅(x−x1)
y-pi=-sqrt3*(x-1)y−π=−√3⋅(x−1)
y=-sqrt3*x+sqrt3+piy=−√3⋅x+√3+π
I hope the explanation is useful...God bless..
