"Let "y=-3x^2sinx
"Let"u=x^2, v=sinx
y=-uv
"Differentiating wrt x on both sides"
(dy)/dx=d/dx(y)=-d/dx(uv)
"By product rule", d/dx(uv)=u(dv)/dx+v(du)/dx
(du)/dx=2x, (dv)/dx=cosx
d/dx(uv)=x^2cosx+(sinx)(2x)
d/dx(uv)=x^2cosx+2xsinx
(dy)/dx=-(x^2cosx+2xsinx)
x=pi/3
y=-3x^2sinx=y=-3(pi/3)^2sin(pi/3)
y=-3(pi)^2/9xx1/2
y=-(pi)^2/6
P-=(x,y)=(pi/3,-(pi)^2/6)
dy/dx=-(x^2cosx+2xsinx)
=-((pi/3)^2cos(pi/3)+2(pi/3)sin(pi/3))
=-(pi^2/9xx1/2+(2pi)/3xxsqrt3/2)
dy/dx=-((pi)^2/18+pi/sqrt3)
"Slope of the tangent at "P-=(pi/3,-(pi)^2/6) "is"
m=dy/dx=-((pi)^2/18+pi/sqrt3)
"Slope of the normal m' is given by " m'=-1/m
m'=1/((pi)^2/18+pi/sqrt3)=1/((sqrt3(pi)^2+18pi)/(18sqrt3))
m'=(18sqrt3)/(sqrt3(pi)^2+18pi)=(18sqrt3)/(pi(sqrt3+18)
m'=(18sqrt3)/(pi(18+sqrt3)
"Equation of the normal passing through the point ", P-=(pi/3,-(pi)^2/6) "and having the slope ", m'=(18sqrt3)/(pi(18+sqrt3) "
"is given by"
(y-(pi)^2/6)/(x-pi/3)=(18sqrt3)/(pi(18+sqrt3)
