How to find dy/dx for parametric equations without eliminating the parameter

Key concept: When $x$ and $y$ both depend on a parameter $t$, the cleanest route is usually to differentiate each part with respect to $t$ and then divide. That avoids messy elimination when the curve is hard to rewrite in Cartesian form.

Step by step

For parametric equations

$x=\phi(t), \qquad y=\psi(t),$

treat both variables as functions of the same parameter $t$.

Step 1: Differentiate with respect to $t$

Compute

$\frac{dx}{dt}=\phi'(t), \qquad \frac{dy}{dt}=\psi'(t).$

Step 2: Use the chain rule

Since $y$ depends on $x$, and $x$ depends on $t$,

$\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}.$

Rearrange to get

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \qquad \text{provided } \frac{dx}{dt}\neq 0.$

Why this is useful

You do not need to eliminate $t$ first. That is especially helpful when the parametric equations come from geometry, motion, or when solving for $y$ in terms of $x$ would be awkward or impossible.

Quick example

If

$x=t^2+1, \qquad y=t^3-4t,$

then

$\frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2-4,$

so

$\frac{dy}{dx}=\frac{3t^2-4}{2t}.$

If needed, you can later rewrite the result in terms of $x$ by using $t=\pm\sqrt{x-1}$.

Pitfall alert

A common mistake is to divide by $\frac{dx}{dt}$ without checking whether it is zero. At values where $\frac{dx}{dt}=0$, the slope may be vertical or undefined, so the formula needs a limit-based check. Another trap is trying to eliminate $t$ too early and making the algebra harder than it needs to be.

Try different conditions

If the question asks for the second derivative, differentiate the first derivative again with respect to $t$:

$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.$

If $x$ is not monotonic in $t$, the same curve can have multiple points with the same $x$-value, so it is often better to keep the parameter form instead of converting to a single $y=f(x)$ equation.

Further reading

parametric equations, chain rule, first derivative