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
Related Maths A Level answers
All answers ▸Find dy/dx such that y=(e^x)(3x+1)^2.
Differentiate 2cos(x)sin(x) with respect to x
Express 2x^2 +8x +7 in the form A(x+B)^2 + C, where A, B and C are constants
C4 June 2014 Q4: Water is flowing into a vase. When the depth of water is h cm, the volume of water V cm^3 is given by V=4πh(h+4). Water flows into the vase at a constant rate of 80π cm^3/s. Find the rate of change of the depth of water in cm/s, when h=6.