Let y = f(x) and its inverse x = g(y) = f^(-1)(y). If p(x) = f'(x), q(x) = f''(x), and r(x) = f'''(x), what are g'(x), g''(x), and g'''(x) in terms of p, q, r, g, and x without derivatives?

Level: Advanced

By the chain rule,
1 = dx/dx = dx/dy * dy/dx.
so
(1) g'(y) = dx/dy = 1/(dy/dx) = 1/p(x).
By the chain rule again,
(2) g''(y) = [g'(y)]' = d(1/p(x))/dy = -1/p^2(x)*p'(x)*(dx/dy) = -q(x)/p^3(x).
Yet another application of chain rule,
(3) g'''(y) = [g''(y)]' = d(-q(x)/p^3(x))'/dy = [-p^3(x)*q'(x)*(dx/dy)+q(x)*3p^2(x)*p'(x)*(dx/dy)]/p^6(x) = (dx/dy)*[-p(x)*r(x)+3q^2(x)]/p^4(x) = (1/p(x))*(-p(x)*r(x)+3q^2(x))/p^4(x) = [-p(x)*r(x)+3q^2(x)]/p^5.
Substituting x = g(y) in (1), (2), and (3), we obtain
g'(y) = 1/p(g(y)),
g''(y) = -q(g(y))/[p(g(y))]^3,
and
g'''(y) = [(-p(g(y))*r(g(y))+3*q^2(g(y))]/[p(g(y))]^5.
Finally, dropping all the y's on both sides of each of these expressions, when it is clear that all the functions are referring to the same variable, we get
g' = 1/p(g), g'' = -q(g)/[p(g)]^3, and g''' = [-p(g)*r(g)+3*q^2(g)]/[p(g)]^5.