differentiation (1 Viewer)

[Dio-Rama]

find the equation of the tangent to the curve y=sin(pi - x)
at the point (pi/6, 1/2), in exact form.

 

[SoulSearcher]

y =sin(pi-x)
y' = -cos(pi-x)
at x = pi/6,
y' = -cos(pi-pi/6)
= -(-cos pi/6)
= (rt3)/2
From here, just use the point-gradient formula using the point (pi/6, 1/2) and the gradient (rt3)/2

 

[zeek]

using the chain rule...

dy/dx = -1.cos(pi - x)

since the co ordinates are (pi/6, 1/2) we sub in the x ordinate to find the gradient...

dy/dx = -1.cos(pi-pi/6)
= -1.-root(3)/2
= sqrt(3)/2
now applying the gradient-intercept formula...

y-1/2 = sqrt(3)/2 . (x - pi/6)
2y - 1= sqrt(3).x - (sqrt(3).pi)/6

and then re arrange the equation to suit u...