If x^2+y^2+ y=4 , then the value of d^2 y d x^2 at the point (-2,0) is

Step-by-step solution

Given, x^2+y^2+ y=4 After differentiating the above equation w . r. t. x we get aligned &2 x+2 y d y d x + y d y d x =0 (1) & 2 x+(2 y+ y) d y d x =0 & d y d x = -2 x 2 y+ y & At (-2,0), ( d y d x )_ (-2,0) = -2 -2 2 0+ 0 & ( d y d x )_ (-2,0) = 4 0+1 & ( d y d x )_ (2,0) =4 (2) aligned Again differentiating equation (1) w. r. t to x , we get 2+2 ( d y d x )^2+2 y d^2 y d x^2 - y ( d y d x )^2+ y d^2 y d x^2 =0 2+(-2 y) ( d y d x )^2+(2 y+ y ) d^2 y d x^2 =0 (2 y+ y ) d^2 y d x^2 =-2-(2- y) ( d y d x )^2 d^2 y d x^