Derivatives of Parametric Functions

This lesson covers how to differentiate parametric functions, where both $x$ and $y$ are given in terms of a parameter such as $t$ or $\theta$. The key idea is to find each derivative with respect to the parameter and then divide them.

The parametric rule

When $x$ and $y$ are both functions of a parameter $t$, the derivative of $y$ with respect to $x$ is

$\frac{dy}{dx} = \frac{\;dy/dt\;}{\;dx/dt\;}.$

The same formula works with any parameter, for example $\theta$ in place of $t$.

Example 1: simple powers

Given $x = at^2$ and $y = 2at$.

Derivatives with respect to $t$:

$\frac{dx}{dt} = 2at, \qquad \frac{dy}{dt} = 2a.$

Combine:

$\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}.$

Example 2: using the quotient rule

Given $x = \dfrac{1 + \log t}{t^2}$ and $y = \dfrac{3 + 2\log t}{t}$.

Differentiate $x$ with the quotient rule:

$\frac{dx}{dt} = \frac{t^2 \cdot \tfrac{1}{t} - (1 + \log t)\cdot 2t}{t^4} = \frac{t - 2t(1 + \log t)}{t^4} = \frac{-t - 2t\log t}{t^4} = -\frac{1 + 2\log t}{t^3}.$

Differentiate $y$ with the quotient rule:

$\frac{dy}{dt} = \frac{t \cdot \tfrac{2}{t} - (3 + 2\log t)\cdot 1}{t^2} = \frac{2 - 3 - 2\log t}{t^2} = -\frac{1 + 2\log t}{t^2}.$

Combine:

$\frac{dy}{dx} = \frac{-\dfrac{1 + 2\log t}{t^2}}{-\dfrac{1 + 2\log t}{t^3}} = \frac{t^3}{t^2} = t.$

The matching factors cancel, leaving $\dfrac{dy}{dx} = t$.

Example 3: a parameter and its reciprocal

Given $x = t + \dfrac{1}{t}$ and $y = t - \dfrac{1}{t}$.

Derivatives with respect to $t$:

$\frac{dx}{dt} = 1 - \frac{1}{t^2}, \qquad \frac{dy}{dt} = 1 + \frac{1}{t^2}.$

Combine and take a common denominator:

$\frac{dy}{dx} = \frac{1 + \tfrac{1}{t^2}}{1 - \tfrac{1}{t^2}} = \frac{\tfrac{t^2 + 1}{t^2}}{\tfrac{t^2 - 1}{t^2}} = \frac{t^2 + 1}{t^2 - 1}.$

Key takeaways

• For parametric curves, $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.
• Use the quotient rule when $x$ or $y$ is itself a fraction in the parameter.
• After dividing, common factors in the parameter often cancel, giving a simple final expression.