Differentiation 3

Given that,
(x^4)(dy/dx)^2 + y^2 = 1 ---------------------(A)
and
y = sin(1/x) ---------------------(B)

In order to show that

(x^4)[(d^2)y/d(x^2)] + 2(x^3)(dy/dx) + y = 0

We need to differentiate (A) by x both sides in order to investigate (d^2)y/d(x^2).

Therefore,

d[(x^4)(dy/dx)^2 + y^2]/dx = d(1)/dx

=> d[(x^4)(dy/dx)^2]/dx + 2y(dy/dx) = 0

=> (x^4)d(dy/dx)^2/dx + (4x^3)(dy/dx)^2 + 2y(dy/dx) = 0

=> (x^4){2(dy/dx)[d(dy/dx)/dx]} + (4x^3)(dy/dx)^2 + 2y(dy/dx) = 0

=> (x^4)2(dy/dx)[(d^2)y/d(x^2)] + (4x^3)(dy/dx)^2 + 2y(dy/dx) = 0

=> 2(x^4)[(d^2)y/d(x^2)](dy/dx) + (4x^3)(dy/dx)^2 + 2y(dy/dx) = 0

=> (x^4)[(d^2)y/d(x^2)](dy/dx) + (2x^3)(dy/dx)^2 + y(dy/dx) = 0

=> (x^4)[(d^2)y/d(x^2)] + (2x^3)(dy/dx) + y = 0