Find the gradient of the curve (x^3)-4(y^2)=12xy at the point P(-8,8)
First of all differentiate the equation of the curve implicitly, giving:
3x2-8y(dy/dx)=12y+12x(dy/dx)
=> (dy/dx)(12x+8y)=3x2-12y
=> dy/dx=(3x2-12y)/(12x+8y)
As dy/dx is the gradient of the curve, if we insert x=-8 and y=8, we will have the gradient of the curve specific to the P location:
dy/dx=[3(-8)2-12(8)]/[12(-8)+8(8)]=-3
FG
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