If x3−2x2y2+5x+y−5=0,y(1)=1 then y′(1)=

x3−2x2y2+5x+y−5=0⇒y1=−3x2−4xy5+51−4x2y⇒y1(1)=−(3−4+5)1−4=43⇒y2=−1−4x2y6x−4y2−8xyy1+3x2−4xy2+5−4x2y1−8xy1−4x2y2⇒y2(1)=−(1−4)6−4−8⋅43+(3−4+5)−4⋅43−8(1−4)2=3−263+4−4039=−23827=−82227