Find an equation of the tangent and normal lines to the curv

Find an equation of the tangent and normal lines to the curve f(x) = x; sin x; at the point ( pi / 6, pi / 12)

Solution

y = x sinx differentiate y\' = xcosx + sinx slope of tangent m = xcosx + sinx at x = pi/6, the slope is y\'(pi/6) = (pi/6)cos(pi/6) + sin(pi/6) => pie/6 sqrt(3)/2+1/2 The equation of tangent with slope 1 and containing point(pi/6,pi/12) is y - pie/12 = pie/6 (sqrt(3)/2)+1/2(x-pi/2) y=0.71+1/2(x-1.57)