# Prove second derivative of $g$ is proportional to $g^2$

From Apostol's Calculus Vol. 1, chapter 6.26, exercise 30:

Let $f(x) = \int_0^x (1+t^3)^{-1/2} dt$.
$a)$ Prove $f$ is strictly monotonic.
$b)$ Let $g$ be the inverse of $f$. Show that the second derivative of $g$ is proportional to $g^2$ and find the constant of proportionality $c$.

The first part is simple, since the derivative is $( 1+x^3 )^{-1/2} > 0$ if $x \ge 0$.

However, I'm having a lot of difficulty with the second part. Since $g' = 1 / f'$, I calculated that $g''(x) = - \frac{3x^2}{2(1+x^3)^{3/2}}$. However, this doesn't seem to get us any closer to finding the constant of proportionality since it doesn't seem at all obvious how to calculate $g$.

I also tried applying the chain rule so that $g''(x) = - f''(x) / f'(x)^2 = - f''(x) g'(x)^2$. Again, this expression doesn't seem to bring us any closer to solving the problem, since we are again left with the problem of calculating $g$, or at least $g^2$, in terms of $f''$ and $g'$.

So how can these equations be manipulated so that we arrive at $g''(x) = cg^2(x)$?

• $g'(x)$ is not $1/f'(x)$; rather $g'(x)=1/f'(g(x))$. ${}\qquad{}$ Aug 29, 2014 at 2:51
• @MichaelHardy oops. I kept thinking in my head $dy/dx = 1/(dx/dy)$ but didn't apply that correctly.
– A.S
Aug 29, 2014 at 2:53
• This is one occasion where working in Leibniz notation might have avoided that confusion. That's what I did in my answer below. Aug 29, 2014 at 3:02
• Au contraire. Working with the Leibniz notation makes it easier not to know/understand what you're doing :) Aug 29, 2014 at 3:27
• In the Leibniz notation like $dy/dx$ it is difficult to think of inverse functions. Because if $y$ is function of $x$ then $x$ is a function of $y$. In the current question when we say that $f, g$ are inverses of each other we tend to think that both $f, g$ are functions of $x$. So while using Leibniz notation we need to be a bit careful about the dependent and independent variables. Aug 31, 2014 at 8:13

$$g'(x) = \frac{1}{f'(g(x))}$$
Since $f'(x) = (1+x^3)^{-1/2}$, $g'(x) = ( 1+g(x)^3 )^{1/2}$. Therefore,
$$g''(x) = \frac{1}{2} ( 1 + g(x)^3 )^{-1/2} \cdot 3 g(x)^2 \cdot g'(x) = \frac{3}{2} g(x)^2$$
It's now clear the constant of proportionality is $3/2$.