What is the second derivative of $f^{-1}(g(x))$?

Question

What is the second derivative of $f^{-1}(g(x))$?

$f^{-1}$ is the inverse of $f$:

• The first derivative of $f^{-1}(x)$ is given by $\frac{1}{f'(f^{-1}(x))}$

• The second derivative of $f^{-1}(x)$ is given by $-\frac{f''(f^{-1}(x))}{[f'(f^{-1}(x))]^3}$

The second derivative of $f(g(x))$ (by the chain rule) is given by $g'(x)^2\cdot f''(g(x))+g''(x) \cdot f'(g(x))$. Substituting the derivatives into this equation, we have that the second derivative of $f^{-1}(g(x))$ is

$$-\frac{g'(x)^2\cdot f''(f^{-1}(g(x)))}{[f'(f^{-1}(g(x)))]^3}+\frac{g''(x)}{f'(f^{-1}(g(x)))}$$

Which is the same as

$$\frac{g''(x)*[f'(f^{-1}(g(x))]^2-g'(x)^2*f''(f^{-1}(g(x)))}{[f'(f^{-1}(g(x)))]^3}$$

Is this correct?

Answer 1

You result seems correct here is an alternative way

Let $h(x) =f^{-1}(g(x)) $ then $$h'(x) =g'(x)(f^{-1})'(g(x)) =g'(x)\frac{1}{f'f^{-1}(g(x))}$$ $$h''(x) =g''(x)\frac{1}{f'f^{-1}(g(x))} +g'(x)\left(\frac{1}{f'f^{-1}(g(x))}\right)'\\=g''(x)\frac{1}{f'f^{-1}(g(x))} -g'(x)\frac{\left(f'(f^{-1}(g(x)))\right)'}{\left(f'f^{-1}(g(x))\right)^2}$$

and $$\left(f'(f^{-1}(g(x)))\right)'=f''(f^{-1}(g(x)))\left(f^{-1}(g(x))\right)' \\= f''(f^{-1}(g(x)))g'(x)\left(f^{-1}\right)'(g(x))\\= \frac{g'(x)f''(f^{-1}(g(x)))}{f'f^{-1}(g(x))}$$

Thus, $$h''(x) =\frac{g''(x)}{f'f^{-1}(g(x))} -g'(x)^2\frac{f''(f^{-1}(g(x)))}{\left(f'f^{-1}(g(x))\right)^3}$$

Answer 2

Your answer is correct.

It's often easier to do it the following way:

If $h(x)=f^{-1}(g(x))$ then $g(x)=f(h(x))$ and thus:
$$\begin{align}g'(x)&=f'(h(x))h'(x)\\ g''(x)&=f''(h(x))h'(x)^2+f'(h(x))h''(x)\end{align}$$ Setting $u(x)=f'(h(x))$ just for shorthand, you then have:

$$\begin{align}h'(x)&=\frac{g'(x)}{u(x)}\\h''(x)&=\frac{g''(x)-f''(h(x))h'(x)^2}{u(x)}\end{align}$$

Substitute the value of $h'(x)$ into the formula for $h''(x)$ (and we get:

$$h''(x)=\frac{g''(x)u^2(x)-f''(h(x))(g')^2(x)}{u^3(x)}$$

Substituting back for $u(x)$ and $h(x)$ gives your same formula, I believe.

Answer 3

Symbolically, (mind you, just symbolically), I would have done it as follows:

$$y = f^{-1}(g(x)) \implies f(y) = g(x) \implies f'(y) \frac{dy}{dx} = g'(x)$$.

Hence, $\frac{dy}{dx} = \frac{g'(x)}{f'(y)}$. Now take second derivative using quotient rule:

$$\frac{d^2y}{dx^2} = \frac{f'(y) \cdot g''(x) - g'(x) \cdot f''(y) \cdot \frac{dy}{dx}}{(f'(y))^2}$$

Then substitute $\frac{dy}{dx} = \frac{g'(x)}{f'(y)}$, simplify, then substitute $y = f^{-1}(g(x))$ wherever there is $y$, and boom! You have the answer! :-)