If x2 + xy + y3 = 1 find the value of y”’ at the point where x = 1.

If x2 + xy + y3 = 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 dr2x 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+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 + 2xs dy_24x2 - 48y-24y -48xy dy x +3y dhx dhr dhx x+3y r +3 72y3-24x2 - 48xy+ 2r x +3 (x2-46y-24+4 x+3y 72y3-24x2 - 48xy+ 2r x+3Find 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(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(

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