A curve has equation x^2 + 2xy – 3y^2 + 16 = 0. Find the coordinates of the points on the curve where dy/dx =0
We would do this by differentiating everything individually so to differentiate x2 we multiply the x2 by the power which is 2 and then take the power down by 1 to make 2x. To differentiate 2xy we would use the product rule. so we would differentiate 2x and y with respect to x to get 2 and dy/dx respectively. The formula of the product rule is where c=ab the differential of c with respect x = (a * db/dx) + ( b* da/dx) meaning it is (2x * dy/dx) + 2y .To differentiate -3y2 ,we do it how we would do it with x which i showed to do earlier but then multiply by dy/dx making -6y * dy/dx we also know that the differential of 16 and 0 is 0 as they are constant. We can then put things in terms of dy/dx to make dy/dx (2x-6y)+2x+2y = 0 this means that dy/dx (2x-6y) = -2x-2y . If dy/dx =0 then the left hand side is 0 meaning -2x-2y =0 meaning that if we divide both sides by -2 we get x+y =0. This can turn into x=-y if you subtract y from both sides meaning if we subsitute -y for x on the original equation we get (-y)2 +2(-y)*y-3y2 +16 = 0 which can be simplified to get -4y2 =-16 if we divide both sides by -4 we get y2 =4 meaning y can be 2 or -2 as from above x=-y we get 2 sets of coordinates (2,-2) and (-2,2)
