Dear Uncle Colin,

What’s the second derivative of $\frac{x^2}{x^2-4}$, and how would you work it out?

- Calculus Obviously Hasn’t Evolved Nicely

Hi, COHEN, and thanks for your message!

There’s an ugly way and a neat way. I’m going to call your expression $y$, so that it has a name.

The ugly way

The ugly way is to do it as it stands using the quotient rule.

$u = x^2$, so $\diff{u}{x} = 2x$; v = $x^2 - 4$, so $\diff{v}{x} = 2x$.

Then $\dydx = \frac{2x\br{x^2 - 4} - 2x(x^2)}{\br{x^2-4}^2}$.

That simplifies a bit to $\frac{-8x}{\br{x^2-4}^2}$.

We can use quotient rule again:

$U = -8x$ so $\diff{U}{x} = -8$; $V = \br{x^2 - 4}^2$, so $\diff{V}{x} = 4x\br{x^2-4}$.

That means $\diffn{2}{y}{x} = \frac{-8\br{x^2-4}^2 + 32x^2\br{x^2-4}}{\br{x^2-4}^4}$.

There’s a factor of $x^2 - 4$ that comes out: $\frac{-8\br{x^2-4} + 32x^2}{\br{x^2-4}^3}$, or $\frac{8\br{4 + 3x^2}}{\br{x^2 - 4}^3}$.

OK. Not too awful, but pretty ugly.

Neater

  • $y = \frac{x^2}{x^2-4} = 1 + \frac{4}{x^2-4}$.
  • $y = 1 + \frac{4}{x^2-4} = 1 + \frac{1}{x-2} - \frac{1}{x+2}$.

That’s a lot easier to differentiate!

  • $\dydx = -\br{x-2}^{-2} + \br{x+2}^{-2}$
  • $\diffn{2}{y}{x} = 2\br{x-2}^{-3} - 2\br{x+2}^{-3}$

That’s a nice answer! If they really want it in fractional form:

  • $\dots = 2\frac{(x+2)^3 - (x-2)^3}{(x-2)^3(x+2)^3}$
  • $\dots = 2\frac{12x^2 + 16}{\br{x^2-4}^3}$
  • $\dots = 8\frac{3x^2 + 4}{\br{x^2-4}^3}$, as before.

Hope that helps!

- Uncle Colin