# Implicit Differentiation of Parametric Equations

I’ve never liked memorizing formulas. I would rather know where they came from or be able to tie it to something I already know. One of my least favorite formulas to remember and explain was the formula for the second derivative of a curve given in parametric form. No longer.

If $y=y\left( t \right)$ and, $x=x\left( t \right)$ then the tradition formula gives $\displaystyle \frac{d}{dx}y\left( t \right)=\frac{dy/dt}{dx/dt}$, and $\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{\frac{d}{dt}\left( \frac{dy}{dx} \right)}{\frac{dx}{dt}}$

It is that last part, where you divide by $\displaystyle \frac{dx}{dt}$, that bothers me. Where did the $\displaystyle \frac{dx}{dt}$ come from?

Then it occurred to me that dividing by $\displaystyle \frac{dx}{dt}$ is the same as multiplying by $\displaystyle \frac{dt}{dx}$

It’s just implicit differentiation!

Since $\displaystyle \frac{dy}{dx}$ is a function of t you must begin by differentiating the first derivative with respect to t. Then treating this as a typical Chain Rule situation and multiplying by $\displaystyle \frac{dt}{dx}$ gives the second derivative. (There is a technical requirement here that given $x=x\left( t \right)$, then $t={{x}^{-1}}\left( x \right)$ exists.)

In fact, if you look at a proof of the formula for the first derivative, that’s what happens there as well: $\displaystyle \frac{d}{dx}y\left( t \right)=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{dy/dt}{dx/dt}$

The reason you to do it this way is that since x is given as a function of t, it may be difficult to solve for t so you can find dt/dx in terms of x. But you don’t have to; just divide by dx/dt which you already know.

Here is an example for both derivatives.

Suppose that $x={{t}^{3}}-3$ and $y=\ln \left( t \right)$

Then $\displaystyle \frac{dy}{dt}=\frac{1}{t}$, and $\displaystyle \frac{dx}{dt}=3{{t}^{2}}$ and $\displaystyle \frac{dt}{dx}=\frac{1}{3{{t}^{2}}}$.

Then $\displaystyle \frac{dy}{dx}=\frac{1}{t}\cdot \frac{dt}{dx}=\frac{1}{t}\cdot \frac{1}{3{{t}^{2}}}=\frac{1}{3{{t}^{3}}}=\frac{1}{3}{{t}^{-3}}$

And $\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \frac{d}{dt}\left( \frac{dy}{dx} \right) \right)\cdot \frac{dt}{dx}=\left( -{{t}^{-4}} \right)\cdot \frac{dt}{dx}=\left( -{{t}^{-4}} \right)\left( \frac{1}{3{{t}^{2}}} \right)=-\frac{1}{3{{t}^{6}}}$

Yes, it’s the same thing as using the traditional formula, but now I’ll never have to worry about forgetting the formula or being unsure how to explain why you do it this way.

Revised: Correction to last equation 5/18/2014

Revised:  2/8/2016

## 7 thoughts on “Implicit Differentiation of Parametric Equations”

1. Ilona DiCosmo on said:

That is very interesting…I don’t like memorizing formulas as well….I’m just thinking, what about test taking…will there be a time to derive it?

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• Lin McMullin on said:

I don’t think there is any “formula” to derive. Given dy/dx as a function of t, you differentiate dy/dx with respect to t and then multiply by dt/dx since you want the second derivative with respect to x (in terms of t). The students have to pay attention to what variable they are using (as always).

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2. Angel Perez on said:

Thank you for this and for all your helpful posts. I can’t reason what happened to the 3; shouldn’t the answer to the second derivative be divided by 3?

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• Lin McMullin on said:

Whoops. Looks like a typo. I have corrected the last equation. Thanks for catching that.

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• Angel Perez on said:

My pleasure. You have been a tremendous help for many years. Thank you.

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3. Katy on said:

Thought provoking and immediately useful post (as always 🙂 )- Thanks! Does this imply that parametric equations can be thought of as compositions?

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• Lin McMullin on said:

In a way, I think they can. If x is a function of t, then t is a function of x, namely the inverse. When you find the second derivative with respect tox of the implicitly defined dy/dx, dividing by dx/dt is the the same as multiplying by dt/dx. At the very least, it is a good way to remember how to find the second derivative which in parametric situations is not just differentiating the first derivative.

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