Derivative of a Function in Parametric Form

Introduction to Parametric Form

In many real-world cases, it’s difficult to express one variable directly in terms of another. For instance, imagine a drone flying in a curved path — its horizontal distance $(x)$ and height $(y)$ both depend on time $(t)$. Instead of writing $y$ as a function of $x$, we express both $x$ and $y$ in terms of the parameter $t$. This is known as a parametric form of a function.

In mathematics, parametric equations describe a curve by introducing a third variable (called the parameter) that defines both $x$ and $y$. For example,
$x = \cos t$ and $y = \sin t$ represent a circle, where $t$ (in radians) controls the position of a point as it moves around the circle.

Why We Use Parameters Instead of Direct $x$–$y$ Relationships

We use parameters when a curve or motion is too complex to describe using a single equation like $y = f(x)$. Parametric equations allow us to study motion and geometry more easily, especially when both $x$ and $y$ depend on another quantity such as time, angle, or speed.

Using parameters gives a clearer picture of how variables change together. It also simplifies the process of differentiation, making it easier to find the derivative of a function in parametric form, where
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

Differentiation

The process of finding the derivative of a function is called differentiation. It helps us understand how one quantity changes with respect to another — for example, how distance changes with time gives us speed.

Let $f$ be a function defined on an open interval $I \subseteq \mathbb{R}$ containing the point $x_0$. If

$\lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$

exists, then $f$ is said to be differentiable at $x_0$, and the derivative of $f$ at $x_0$, denoted by $f'(x_0)$, is given by

$f'(x_0) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$

For all $x$ where this limit exists, the derivative becomes a function of $x$ given by

$f'(x) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$

The derivative represents the instantaneous rate of change of $f(x)$ with respect to $x$, and it is the fundamental concept in calculus used to study slopes, velocity, and rates of change in various applications.

Standard Notations for Derivatives

There are several notations used to represent the derivative of $y = f(x)$, such as
$f'(x)$, $\frac{dy}{dx}$, $y'$, $\frac{d}{dx}[f(x)]$, $D_x[y]$, $Dy$, or $y_1$.

Here, $\frac{d}{dx}$ or $D$ is known as the differential operator, which indicates differentiation with respect to $x$.

Differentiation of Some Basic Functions

These are the fundamental rules that form the base for all differentiation formulas in calculus:

1. $\frac{d}{dx}(\text{constant}) = 0$

2. $\frac{d}{dx}(x^n) = n x^{n-1}$

3. $\frac{d}{dx}(a^x) = a^x \log_e a$

4. $\frac{d}{dx}(e^x) = e^x$

5. $\frac{d}{dx}(\log_e |x|) = \frac{1}{x}, \ x \neq 0$

6. $\frac{d}{dx}(\log_a |x|) = \frac{1}{x \log_e a}, \ x \neq 0$

7. $\frac{d}{dx}(\sin x) = \cos x$

8. $\frac{d}{dx}(\cos x) = -\sin x$

9. $\frac{d}{dx}(\tan x) = \sec^2 x$

10. $\frac{d}{dx}(\cot x) = -\csc^2 x$

11. $\frac{d}{dx}(\sec x) = \sec x \tan x$

12. $\frac{d}{dx}(\csc x) = -\csc x \cot x$

These basic derivative rules are essential for solving higher-level problems in calculus and parametric differentiation.

Differentiation of Function in Parametric Form

Parametric differentiation is used when both the dependent variable $y$ and the independent variable $x$ are expressed in terms of another variable, usually $t$.

In such cases, $x = g(t)$ and $y = f(t)$ are called parametric equations, and $t$ is known as the parameter.

To find $\frac{dy}{dx}$ in parametric form, we apply the chain rule. Since both $x$ and $y$ are functions of $t$, we first differentiate each with respect to $t$ and then take their ratio:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{f'(t)}{g'(t)}$

This formula helps us find how $y$ changes with respect to $x$ without having to eliminate the parameter $t$, which is often difficult or impossible for complex curves.

Example of Differentiation in Parametric Form

Let
$x = a(1 - \cos \theta)$ and $y = a(\theta - \sin \theta)$.
To find $\frac{dy}{dx}$, differentiate each with respect to $\theta$:

$\frac{dx}{d\theta} = a(\sin \theta)$ and $\frac{dy}{d\theta} = a(1 - \cos \theta)$

Now,

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{1 - \cos \theta}{\sin \theta}$

This result shows the slope of the curve represented by these parametric equations at any point defined by $\theta$.

Second Derivative in Parametric Form

Sometimes, we also need the second derivative $\frac{d^2y}{dx^2}$ to analyze the curvature or concavity of the graph. Using the chain rule,

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

For example, consider the circle defined by $x = \cos t$ and $y = \sin t$.
Here, $\frac{dy}{dx} = -\cot t$. Differentiating again with respect to $t$:

$\frac{d}{dt}(-\cot t) = \csc^2 t$

Since $\frac{dx}{dt} = -\sin t$,

$\frac{d^2y}{dx^2} = \frac{\csc^2 t}{-\sin t} = -\csc^3 t$

This process of second-order differentiation in parametric form helps in understanding the shape, curvature, and turning points of a curve defined by parametric equations.

Solved Examples Based on Differentiation of Parametric Form:

Example 1: If $x = 2 \sin \theta - \sin 2\theta$ and $y = 2 \cos \theta - \cos 2\theta$, $\theta \in [0, 2\pi]$, then find $\frac{d^2 y}{dx^2}$ at $\theta = \pi$.
[JEE Main 2020]

1. $\frac{3}{8}$
2. $\frac{3}{4}$
3. $\frac{3}{2}$
4. $-\frac{3}{4}$

Solution

$\frac{dx}{d\theta} = 2 \cos \theta - 2 \cos 2\theta$

$\frac{dy}{d\theta} = -2 \sin \theta + 2 \sin 2\theta$

$\therefore \frac{dy}{dx} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}$

$= \frac{2 \sin \frac{\theta}{2} \cos \frac{3\theta}{2}}{2 \sin \frac{\theta}{2} \sin \frac{3\theta}{2}} = \cot \frac{3\theta}{2}$

Now, $\frac{d^2 y}{dx^2} = \frac{-3}{2} \csc^2 \frac{3\theta}{2} \cdot \frac{d\theta}{dx}$

$\frac{d^2 y}{dx^2} = \frac{-\frac{3}{2} \csc^2 \frac{3\theta}{2}}{2(\cos \theta - \cos 2\theta)}$

$\left.\frac{d^2 y}{dx^2}\right|_{\theta = \pi} = -\frac{3}{4(-1 - 1)} = \frac{3}{8}$

Hence, the correct answer is Option (1).

Example 2: Let $y = \sin^3 t$ and $x = 1 + \cos^3 t$. Find $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$.

1. $2$
2. $-2$
3. $-1$
4. $1$

Solution

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{d(\sin^3 t)}{dt}}{\frac{d(1 + \cos^3 t)}{dt}} = \frac{3 \sin^2 t \cos t}{3 \cos^2 t (-\sin t)}$

$\Rightarrow \frac{dy}{dx} = \frac{3 \sin^2 t \cos t}{-3 \sin t \cos^2 t} = -\tan t$

$\left.\frac{dy}{dx}\right|_{t = \frac{\pi}{4}} = -1$

Hence, the correct answer is Option (3).

Example 3: For $a > 0$, $t \in \left[0, \frac{\pi}{2}\right]$, let $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$. Then find $1 + \left[\frac{dy}{dx}\right]^2$.
[JEE Main 2013]

1. $\frac{x^2}{y^2}$
2. $\frac{y^2}{x^2}$
3. $\frac{x^2 + y^2}{y^2}$
4. $\frac{x^2 + y^2}{x^2}$

Solution

Let $x = \sqrt{a^{\sin^{-1} t}}$

$\Rightarrow x^2 = a^{\sin^{-1} t}$

$\Rightarrow 2 \log x = \sin^{-1} t \cdot \log a$

$\Rightarrow \frac{2}{x} = \frac{\log a}{\sqrt{1 - t^2}} \cdot \frac{dt}{dx}$

$\Rightarrow \frac{2 \sqrt{1 - t^2}}{x \log a} = \frac{dt}{dx}$

Now, let $y = \sqrt{a^{\cos^{-1} t}}$

$\Rightarrow 2 \log y = \cos^{-1} t \cdot \log a$

$\Rightarrow \frac{2}{y} \cdot \frac{dy}{dx} = \frac{-\log a}{\sqrt{1 - t^2}} \cdot \frac{dt}{dx}$

$\Rightarrow \frac{2}{y} \cdot \frac{dy}{dx} = \frac{-\log a}{\sqrt{1 - t^2}} \times \frac{2 \sqrt{1 - t^2}}{x \log a}$

$\Rightarrow \frac{dy}{dx} = -\frac{y}{x}$

Hence,

$1 + \left(\frac{dy}{dx}\right)^2 = 1 + \left(\frac{-y}{x}\right)^2 = \frac{x^2 + y^2}{x^2}$

Therefore, the correct answer is Option (4).

Example 4: Let $y = a e^t$, $x = b \sin(e^t)$. Find $\frac{dy}{dx}$.

1. $\frac{a}{b} \cos(e^t)$
2. $\frac{a}{b} \sec(e^t)$
3. $\frac{a}{b}$
4. $\frac{b}{a} \cos(e^t)$

Solution

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

$= \frac{\frac{d(a e^t)}{dt}}{\frac{d[b \sin(e^t)]}{dt}}$

$= \frac{a e^t}{b \cos(e^t) \cdot e^t}$

$= \frac{a}{b} \sec(e^t)$

Hence, the correct answer is Option (2).

Example 5: Let $y = \sin t$ and $x = \cos t$. Find $\frac{dy}{dx}$.

1. $\tan t$
2. $-\cot t$
3. $-\tan t$
4. $\cot t$

Solution

Given $y = \sin t$ and $x = \cos t$

$\frac{dy}{dt} = \cos t$

$\frac{dx}{dt} = -\sin t$

$\Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos t}{-\sin t} = -\cot t$

Also, $\frac{dy}{dx} = \frac{-x}{y}$

Hence, the correct answer is Option (2).

List of topics related to the Differentiation of Parametric form

Explore all key concepts under the differentiation of parametric equations. This section helps you understand every topic relevant for Class 12 and JEE-level calculus preparation.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Continuity and Discontinuity

Continuity of Composite Function

Continuity And Differentiability

Differentiability of Composite Function

NCERT Resources

Find complete NCERT-based resources like exemplar problems, step-by-step solutions, and concise revision notes on differentiation of parametric form. This section is ideal for building strong conceptual clarity aligned with board and competitive exams.

NCERT Class 12 Maths Notes for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Solutions for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Exemplar Solutions for Chapter 5 - Continuity and Differentiability

Practice Questions based on Differentiation of Parametric form

Get a collection of important practice problems on parametric differentiation ranging from basic to advanced levels. This section includes JEE, CUET, and board-style questions designed to test application and problem-solving skills.

Differentiation Of Function In Parametric Form- Practice Question MCQ

We have shared below the links to practice questions on the related topics of differentiation of parametric functions: