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 \displaystyle y=y(t) and, \displaystyle x=x(t) then the traditional formulas give

\displaystyle \frac{{dy}}{{dx}}=\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 \displaystyle x=x(t), then its inverse \displaystyle 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 happening there as well:

\displaystyle \frac{d}{{dx}}y(t)=\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 \displaystyle x={{t}^{3}}-3 and \displaystyle 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}}}

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 \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. Originally posted May 5, 2014.

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