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 $g''(x) = cg^2(x)$?

  • $\begingroup$ $g'(x)$ is not $1/f'(x)$; rather $g'(x)=1/f'(g(x))$. ${}\qquad{}$ $\endgroup$ Aug 29, 2014 at 2:51
  • $\begingroup$ @MichaelHardy oops. I kept thinking in my head $dy/dx = 1/(dx/dy)$ but didn't apply that correctly. $\endgroup$
    – A.S
    Aug 29, 2014 at 2:53
  • $\begingroup$ This is one occasion where working in Leibniz notation might have avoided that confusion. That's what I did in my answer below. $\endgroup$ Aug 29, 2014 at 3:02
  • 1
    $\begingroup$ Au contraire. Working with the Leibniz notation makes it easier not to know/understand what you're doing :) $\endgroup$ Aug 29, 2014 at 3:27
  • $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    Aug 31, 2014 at 8:13

2 Answers 2


I made a mistake above. The correct expression is:

$$g'(x) = \frac{1}{f'(g(x))}$$

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 $3/2$.


\begin{align} y & = \int_0^x (1+t^3)^{-1/2} \, dt, \\[10pt] \frac{dy}{dx} & = (1+x^3)^{-1/2}, \\[10pt] \frac{dx}{dy} & = \frac{1}{(1+x^3)^{-1/2}} = (1+x^3)^{1/2} = \text{some function of }y. \\[10pt] \frac{d^2x}{dy^2} & = \frac 1 2 (1+x^3)^{-1/2}\cdot3x^2\cdot\frac{dx}{dy} = \frac 1 2 (1+x^3)^{-1/2}\cdot 3x^2\cdot(1+x^3)^{1/2} = \frac 3 2x^2, \\[10pt] g''(y) & = \frac 3 2 (g(y))^2. \end{align}


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