Paragraph: Consider the functions defined implicitly by the equation y^3-3 y+x=0 on various intervals in the real line. If x (- ,-2) (2, ) , the equation implicitly defines a unique real valued differentiable function y=f(x) . If x (-2,2) , the equation implicitly defines a unique real valued differentiable function y=g(x) , satisfying g(0)=0 . Question: If f(-10 2 )=2 2 , then f^ (-10 2 ) is equal to

Step-by-step solution

Given, y^3-3 y+x=0 3 y^2 d y d x -3 d y d x +1=0 3 y^2 ( d^2 y d x^2 )+6 y ( d y d x )^2-3 d^2 y d x^2 =0 On substituting x=-10 2 , y=2 2 in Eq. (i), we get 3(2 2 )^2 d y d x -3 d y d x +1=0 d y d x = -1 21 Again on substituting x=-10 2 , y=2 2 in Eq. (ii), we get aligned & 3(2 2 )^2 d^2 y d x^2 +6(2 2 ) ( -1 21 )^2-3 d^2 y d x^2 =0 & 21 d^2 y d x^2 =- 12 2 (21)^2 d^2 y d x^2 =- 12 2 (21)^3 =- 4 2 7^3 3^2 aligned