# Ask Uncle Colin: Parametric Second Derivatives

Dear Uncle Colin,

I have a pair of parametric equations giving $x$ and $y$ each as a function of $t$. I'm happy with the first derivative being $\diff{y}{t} \div \diff{x}{t}$, but I struggle to find the second derivative. How would I do that?

– Can't Handle An Infinitesimal Nuance

Hi, CHAIN, and thanks for your message!

This always used to trip me up as well – it stood to reason that if the first derivative was $\diff yt \div \diff xt$, then the second derivative should be $\diffn 2yt \div \diffn 2xt$. If only life were so simple.

Instead, the thing to do is to treat your first derivative, $\diff yx$ as a function of $t$ – let's call it $z(t)$ and differentiate it with respect to $x$. That gives you $\diff zx = \diff zt \div \diff xt$.1. However, we still need to figure out $\dot z$, which we can do using the quotient rule: because $z = \frac{\dot y}{\dot x}$, we have $\dot z = \frac{\dot x \ddot y – \ddot x \dot y}{\dot x^2}$.

This finally gives you $\diffn 2yx = \dot z = \frac{\dot x \ddot y – \ddot x \dot y}{\dot x^3}$ for the second derivative.

Hope that helps!

— Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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1. I'll switch to Newton notation here; this is one of the few places it comes out more neatly. []