Question

How do you use implicit differentiation to find `(d^2y)/(dx^2)` given `x^2-y^2=1`?

Answer 1

`y''(x)= pm1/(x^2-1)^(3//2)`

Explanation:

Calling `f(x,y(x))=x^2-y(x)^2-1=0`

`(df)/(dx)=2 x - 2 y(x)y'(x)=0` and

`(d^2f)/(dx^2)=2-2y(x)y''(x)-2y'(x)^2=0`

so

`y''(x)=(y(x)^2-x^2)/(y(x))^3 = pm1/(x^2-1)^(3//2)`