#"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)#
