If x^{2} +
xy + y^{3} = 1
find the value of y”’ at the point where x =
1

## Answer

Consider the equation x +xy+y-1 Need to find by using implicit differentiation. Differentiate with respect to x, get dv dh dr dh Treat y as a function of x and use chain rule, get dy 2x(1)+x+y(1)+3y dr dr Since _ ( Constant )-0 and (x” ) = nxn-1 (2x + y)-(x+3y*) =0 dr d 2rty) -(2x+v -(2x+y) x+3y

2x y Again differentiate v- with respect to x by treating y as constant. (thy x +3 du d d-(2x+y) ахах Since (x+3y2) [-(2x + y)]-[-(2x+y)] (x+3y2) x +3 r +3 ( d (x+3y) r +3 t + ах 2x+ y x +3y dx (x+3y’ (x+3y

-(2x+ y) x+3y x+3y Again differentiate v” with respect to x by treating x+3y y as constant. (x+3) x+3 dr 18y4-24x2y-24m2 + 2×2 + 2xy-(x + 3y2 dhx

s dy_24x2 – 48y-24y -48xy dy x +3y dhx dhr dhx x+3y r +3 72y3-24×2 – 48xy+ 2r x +3 (x2-46y-24+4 x+3y 72y3-24×2 – 48xy+ 2r x+3y -(2r+y) x +3

Find the value of y when x-1. 12 +1(x)-У3-1 So the only real number which satisfies this equation is y-0. Now substitute x,v 0 in the equation v” and simplify -(2x+y) x +3 72y -24×2 -48xy+2x – 46 y 24y’ +4x -(2x+ y x+3y x+3y

(1+3(0′!(-72(0-24(1-48(1)(0)+2(1)(噐 1) (1+3(0) (-72(0 -240-48(0 (20046(0)-24(0+40) 18(0-240 (0)-24)(+2(0)+20)(0)(3)(1+3(01+6(0) 1+3(0) 1+3(0) [(-24 +2)(-2) + 4(1)]-미(3)(1)2(1-0) [H1 = 42