Ask Uncle Colin: A second derivative

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 $\frac{du}{dx}=2x$; v = $x^2-4$, so $\frac{dv}{dx}=2x$.

Then $\frac{dy}{dx}=\frac{2x(x^2-4)-2x(x^2)}{{(x^2-4)}^2}$.

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

We can use quotient rule again:

$U=-8x$ so $\frac{dU}{dx}=-8$; $V={(x^2-4)}^2$, so $\frac{dV}{dx}=4x(x^2-4)$.

That means $\frac{d^{2}y}{d{x}^2}=\frac{-8{(x^2-4)}^2+32x^2(x^2-4)}{{(x^2-4)}^4}$.

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

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!

• $\frac{dy}{dx}=-{(x-2)}^{-2}+{(x+2)}^{-2}$
• $\frac{d^{2}y}{d{x}^2}=2{(x-2)}^{-3}-2{(x+2)}^{-3}$

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

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

Hope that helps!

- Uncle Colin