Dear Uncle Colin,

I’m told that $x^2 \tan(y) = 9$, for $0<y< \piby 2$. I have to show that $\dydx = \frac{-18x}{x^4+81}$, and that there’s a point of inflection at $x = 27^{1/4}$. Where do I even start!?

- I Might Plot Loci… I Can’t, It’s Tricky

Hi, IMPLICIT, and thanks for your message!

Plotting this does seem a bit over the top, now you mention it! You could probably figure out a few points – when $x = \pm 1$, $y = \piby 4$, for example; as $y$ approaches 0, $x^2$ increases without bound (so $x$ goes to infinity in both directions), and when $y$ approaches $\piby 2$, $x^2$ approaches 0$. I can picture that! It looks a bit like the Gaussian.

But that’s beside the point! We need to differentiate this, and doing that implicitly is probably for the best.

First part

The left hand side is a product, so we need the product rule: we get $(2x)(\tan(y)) + \br{x^2}\br{\sec^2(y) \dydx} = 0$.

Now, how to manipulate that into what they have? It makes sense to eliminate the $y$s, since there are none of those in the final answer: $\tan(y) = \frac{9}{x^2}$ and we can use the fact that $\sec^2(y) = 1 + \tan^2(y)$ to say $\sec^2(y) = \frac{x^4 + 81}{x^4}$. That has things in common with our answer!

Where are we now, then? $(2x)\br{\frac{9}{x^2}} + \br{x^2}\frac{x^4 + 81}{x^4}\dydx = 0$.

We know that $x \ne 0$, because $y$ never reaches $\piby 4$, so we can mulitply through by $x^2$, cancel and consolidate a little: $18x + \frac{x^4 + 81}\dydx = 0$.

Aha! We’re pretty much there – just shuffle and divide to get $\dydx = \frac{-18x}{x^4 + 81}$ as required.

Second part

For a point of inflection, the second derivative needs to be zero. We can differentiate using the quotient rule (for a moment I wondered about differentiating implicitly again, but haha nope.)

We’ve got $u = -18x$ and $v = x^4 + 81$, so $u’ = -18$ and $v’ = 4x^3$.

$\diffn{2}{y}{x} = \frac{vu’ - uv’}{v^2} = \frac{-18\br{x^4+81} + 72x^4}{x^8}$.

We’re interested in where the top vanishes, i.e., where $-18\br{x^4+81} + 72x^4 =0$.

Taking out $18$ gives $18\br{3x^4 - 81}=0$, so $x^4 = 27$ as required.

Hope that helps!

- Uncle Colin