Derivative of a Function in Parametric Form

Parametric Differentiation is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of differentiation have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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This Story also Contains

1. Differentiation
2. Differentiation of some basic functions
3. Differentiation of Function in Parametric Form
4. Solved Examples Based on Differentiation of Parametric Form:

In this article, we will cover the concept of the Parametric Differentiation. This concept falls under the broader category of Calculus, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last five years of the JEE Main exam (from 2013 to 2023), a total of three questions have been asked on this concept, including one in 2013, one in 2020, and one in 2022.

Differentiation

The process of finding the derivative is called differentiation. Let $f$ be defined on an open interval $I\subseteq$ containing the point $x_0$, and suppose that $\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x_0+\mathrm{\Delta }x)-f\left(x_0\right)}{\mathrm{\Delta }x}$ exists. Then $f$ is said to be differentiable at $x_0$ and the derivative of $f$ at $x_0$, denoted by $f^{\prime }\left(x_0\right)$, is given by

$f^{\prime }\left(x_0\right)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x_0+\mathrm{\Delta }x)-f\left(x_0\right)}{\mathrm{\Delta }x}$

For all $x$ for which this limit exists,
$f^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}$ is a function of $x$.

In addition to $f^{\prime }(x)$, other notations are used to denote the derivative of $y=f(x)$. The most common notations are $f^{\prime }(x),\frac{dy}{dx},y^{\prime },\frac{d}{dx}[f(x)],D_x[y]$ or $Dy$ or $y_1$. Here $\frac{d}{dx}$ or $D$ is the differential operator.

Differentiation of some basic functions

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

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

3. $\frac{d}{dx}\left({\mathbf{a}}^{\mathrm{x}}\right)={\mathbf{a}}^{\mathrm{x}}{\log }_{\mathrm{e}}\mathbf{a}$

4. $\frac{d}{dx}\left({\mathrm{e}}^{\mathrm{x}}\right)={\mathrm{e}}^{\mathrm{x}}{\log }_{\mathrm{e}}\mathrm{e}={\mathrm{e}}^{\mathrm{x}}$

5. $\frac{d}{dx}({\log }_{\mathrm{e}}|\mathbf{x}|)=\frac{\mathbf{1}}{\mathbf{x}},\mathbf{x}\ne 0$

6. $\frac{d}{dx}({\log }_{\mathbf{a}}|\mathbf{x}|)=\frac{1}{\mathbf{x}{\log }_{\mathrm{e}}\mathbf{a}},\mathbf{x}\ne 0$

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

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

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

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

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

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

Differentiation of Function in Parametric Form

Parametric differentiation is the process of finding the derivative of the equation in which the dependent variable $y$ and independent variable $x$ are equated to another variable $t$. To find the derivative of $y$ for $x$ we use the chain rule.

Sometimes, $x$ and $y$ are given as functions of a single variable, i.e., $x=g(t)$ and $y=f(t)$ are two functions and $t$ is a variable. In such cases, $x$ and $y$ are called parametric functions or parametric equations and $t$ is called the parameter.

To find $\frac{dy}{dx}$ in such cases, first find the relationship between $x$ and $y$ by eliminating the parameter $t$ and then differentiate concerning $t$.

We have $x=f(t)$ and $y=f(t)$, in this case, we will differentiate both functions separately. We will first differentiate $x$ and $y$ concerning '$t$ separately. On differentiating $x$ for ' $t$ ' we get $\frac{dx}{dt}$ and on differentiating $y$ by ' $t$ ' we get $\frac{dy}{dt}$.

But sometimes it is not possible to eliminate $t$, then in that case use

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{f^{\prime }(t)}{g^{\prime }(t)}$

For example

If $x=a(1-\cos \theta )$ and $y=a(\theta -\sin \theta )$, then $dy/dx$ is
Solution.

$\begin{array}{rl}& \text{ Given }\mathrm{x}=\mathrm{a}(1-\cos \theta )\text{ and }\mathrm{y}=\mathrm{a}(\theta +\sin \theta )\\ & \Rightarrow \frac{dx}{d\theta }=a(\sin \theta )\text{ and }\frac{dy}{d\theta }=a(1+\cos \theta )\\ & \Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{1+\cos \theta }{\sin \theta }\end{array}$
Using the circle example, we differentiate $(-\cot (t))$ concerning

$\begin{array}{rl}& (t):[\frac{d}{dt}(-\cot (t))={\csc }^2(t)]\text{ Since }(\frac{dx}{dt}=-\sin (t)),\text{ wefind }\left(\frac{d^{2}y}{d{x}^2}\right):[\frac{d^{2}y}{d{x}^2}\\ & =\frac{{\csc }^2(t)}{-\sin (t)}=-\csc (t)]\end{array}$

Recommended Video Based on Differentiation of Parametric Form

Solved Examples Based on Differentiation of Parametric Form:

Example 1: If $x=2\sin \theta -\sin 2\theta$ and $y=2\cos \theta -\cos 2\theta ,\theta \equiv [0,2\pi {]}_{\text{, then }}$ $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}$ at $\theta =\pi$ is :
[JEE Main 2020]
1) $\frac{3}{8}$
2) $\frac{3}{4}$
3) $\frac{3}{2}$
4) $-\frac{3}{4}$

Solution

$\begin{array}{c}\frac{dx}{d\theta }=2\cos \theta -2\cos 2\theta \\ \frac{dy}{d\theta }=-2\sin \theta +2\sin 2\theta \\ \therefore \frac{dy}{dx}=\frac{\sin 2\theta -\sin \theta }{\cos \theta -\cos 2\theta }\\ =\frac{2\sin \frac{\theta }{2}\cdot \cos \frac{3\theta }{2}}{2\sin \frac{\theta }{2}\sin \frac{3\theta }{2}}=\cot \frac{3\theta }{2}\\ \frac{d^{2}y}{d{x}^2}=\frac{-3}{2}{\csc }^2\frac{3\theta }{2}\cdot \frac{d\theta }{dx}\\ \frac{d^{2}y}{d{x}^2}=\frac{-\frac{3}{2}{\csc }^2\frac{3\theta }{2}}{2(\cos \theta -\cos 2\theta )}\\ {\frac{d^{2}y}{d{x}^2}|}_{\theta =\pi }=-\frac{3}{4(-1-1)}=\frac{3}{8}\end{array}$

Hence, the answer is the option 1.

Example 2: Let $y={\sin }^{3}t$ and $x=1+{\cos }^{3}t$, then $\frac{dy}{dx}$ at $t=\frac{\pi }{4}$ is
1) $2$
2) $-2$
3) $-1$
4) $1$

Solution:
$\begin{array}{rl}& \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{d({\sin }^{3}t)}{dt}}{\frac{d(1+{\cos }^{3}t)}{dt}}=\frac{3{\sin }^{2}t\cdot \cos t}{0+(3{\cos }^{2}t)(-\sin t)}\\ & \Rightarrow \frac{dy}{dx}=\frac{3{\sin }^{2}t\cos t}{-3\sin t{\cos }^{2}t}=-\tan t\\ & {\frac{dy}{dx}|}_{t=\frac{\pi }{4}}=-1\end{array}$
Hence, the answer is the option 3.

Example 3: For $a>0,tϵ{[0,\frac{\pi }{2}]}_{\text{let }x}\sqrt{a^{{\sin }^{-1}t}}$ and $y=\sqrt{a^{{\cos }^{-1}t}}$. Then, $1+{\left[\frac{dy}{dx}\right]}^2$ equals to :
[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:
$\begin{array}{rl}& \text{ Let }x=\sqrt{a^{{\sin }^{-1}t}}\\ & \Rightarrow x^2=a^{{\sin }^{-1}t}\\ & \Rightarrow 2\log x={\sin }^{-1}t\cdot \log a\\ & \Rightarrow \frac{2}{x}=\frac{\log a}{\sqrt{1-t^2}}\cdot \frac{dt}{dx}\\ & \Rightarrow \frac{2\sqrt{1-t^2}}{x\log a}=\frac{dt}{dx}\end{array}$
Now, let $y=\sqrt{a^{{\cos }^{-1}t}}$

$\begin{array}{rl}& \Rightarrow 2\log y={\cos }^{-1}t\cdot \log a\\ & \Rightarrow \frac{2}{y}\cdot \frac{dy}{dx}=\frac{-\log a}{\sqrt{1-t^2}}\cdot \frac{dt}{dx}\\ & \Rightarrow \frac{2}{y}\cdot \frac{dy}{dx}=\frac{-\log a}{\sqrt{1-t^2}}\times \frac{2\sqrt{1-t^2}}{x\log a}\\ & \Rightarrow \frac{dy}{dx}=-\frac{y}{x}\end{array}$

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

Example 4: Let $y=a{e}^t,x=b\sin \left(e^t\right)$, then $\frac{dy}{dx}$ equals
1) $\frac{a}{b}\cos \left(e^t\right)$
2) $\frac{a}{b}\sec \left(e^t\right)$
3) $\frac{a}{b}$
4) $\frac{b}{a}\cos \left(e^t\right)$

Solution:

$\begin{array}{rl}& \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ & =\frac{\frac{d}{dt}\left(a{e}^t\right)}{\frac{d}{dt}b\sin \left(e^t\right)}\\ & =\frac{a{e}^t}{b\cos \left(e^t\right)\cdot e^t}\\ & =\frac{a}{b}\sec \left(e^t\right)\end{array}$

Hence, the answer is the option 2.

Example 5:
Let $y=\sin t$ and $x=\cos t$, then $\mathrm{dy}/\mathrm{d}x$ equals
1) $\tan t$
2) $-\cot t$
3) $-\tan t$
4) $\cot t$

Solution:
As we have learned
Let $y=\sin t$
$x=$ cost
$\frac{dy}{dt}=\cos t$
$\frac{dx}{dt}=-\sin t$
$\Rightarrow \frac{dy}{dx}=-\frac{\cos t}{\sin t}$
$=\frac{-x}{y}$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\frac{\frac{d(\sin t)}{dt}}{\frac{d(\cos t)}{dx}}=\frac{\cos t}{-\sin t}=-\cot (t)$

Hence, the answer is the option 2.