Differentiation: Definition, Rule, Formula, Examples

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.

In this article, we will cover the concept of 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 ten years of the JEE Main exam (from 2013 to 2023), a total of seven questions have been asked on this concept, including one in 2018, two in 2019, one in 2021, and three in 2023.

Differentiation

The rate of change of a quantity $y$ concerning another quantity $x$ is called the derivative or differential coefficient of $y$ concerning $x$. Geometrically, the Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point.

Let $y=f(x)$ be a function of $x$. Then, the rate of change of " $y$ " per unit change in " $x$ " is given by: $d y$ $/ \mathrm{dx}$

If the function $f(x)$ undergoes an infinitesimal change of 'h' near any point ' $x$ ', then the derivative of the function is defined as

$
\lim\limits _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
$

Properties of Differentiation:

When a function is denoted as $y=f(x)$, the derivative is indicated by the following notations.
1. $\mathrm{D}(\mathrm{y})$ or $\mathrm{D}[\mathrm{f}(\mathrm{x})]$ is called Euler's notation.
2. $\mathrm{dy} / \mathrm{dx}$ is called Leibniz's notation.
3. $F^{\prime}(x)$ is called Lagrange's notation.

The meaning of differentiation is the process of determining the derivative of a function at any point.

Consider any function, $y=f(x)$
Let $P(x, y)$ and $Q(x+\Delta x, f(x+\Delta x))$ be two points on the curve

The slope of chord $P Q=$

$
\frac{f(x+\Delta x)-f(x)}{(x+\Delta x)-x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}
$

(Using two points lying on the straight line $P Q$ )

Now as $\Delta x \rightarrow 0$, the point $Q$ moves infinitesimally close to point $P$, and chord $P Q$ becomes tangent to the curve at point $P$.

So slope of chord PQ becomes the slope of the tangent to the curve at point $P$
So the slope of the tangent at $\mathrm{P}=\lim\limits _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
This gives the slope of the tangent at P . This value is also known as the derivative or differentiation of function $f(x)$ concerning $x$.

It is also denoted by $\frac{d f(x)}{d x}$ or $\frac{d y}{d x}$ or $f^{\prime}(x)$ or $D(f(x))$

If we replace $\Delta \mathrm{x}$ with h , then we can write

$
\begin{aligned}
\frac{\mathrm{dy}}{\mathrm{dx}}=\lim\limits _{\Delta \mathrm{x} \rightarrow 0} & \frac{\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{x})}{\Delta \mathrm{x}} \\
& =\lim\limits _{h \rightarrow 0} \frac{\mathrm{f}(\mathrm{x}+h)-\mathrm{f}(\mathrm{x})}{h}
\end{aligned}
$
This method to find the derivative of $f(x)$ is also known as the "First Principle of derivative" or "a-b initio method"

Rules of Differentiation (Sum/Difference/Product)

Let $f(x)$ and $g(x)$ be differentiable functions and $k$ be a constant. Then each of the following rules holds:

Sum Rule
The derivative of the sum of a function $f$ and a function $g$ is the same as the sum of the derivative of $f$ and the derivative of g .

$
\frac{d}{d x}(f(x)+g(x))=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))
$

Difference Rule
The derivative of the difference of a function $f$ and a function $g$ is the same as the difference of the derivative of $f$ and the derivative of $g$.

$
\frac{d}{d x}(f(x)-g(x))=\frac{d}{d x}(f(x))-\frac{d}{d x}(g(x))
$

Product rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then,

$
\frac{d}{d x}(f(x) g(x))=g(x) \cdot \frac{d}{d x}(f(x))+f(x) \cdot \frac{d}{d x}(g(x))
$

Solved Examples Based on Differentiation:

Example 1: If $x^2+y^2+\sin y=4$, then the value of $\frac{d^2 y}{d x^2}$ at the point $(-2,0)$ is :
[JEE Main 2018]
1) -34
2) -32
3) 4
4) -2

Solution:

$
x^2+y^2+\sin y=4
$
On Differentiating, we get

$
\begin{aligned}
& 2 x+2 y \frac{d y}{d x}+\cos y \frac{d y}{d x}=0 \\
& \frac{\mathrm{dy}}{\mathrm{d} x}=-\left(\frac{2 x}{2 y+\cos y}\right)
\end{aligned}
$

$\begin{aligned} & \text { at }(-2,0) \\ & \frac{\mathrm{dy}}{\mathrm{d} x}=4 \\ & \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=-\left.\frac{(2 y+\cos y)(2)-(2 y+\cos y)(2 x)}{(2 y+\cos y)^2}\right|_{(-2,0)} \\ & \text { at }(-2,0) \\ & \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=-34\end{aligned}$

Hence, the answer is the option 1.

Example 2: If $\int_0^x f(t) d t=x^2+\int_x^1 t^2 f(t) d t{ }_{\text {, then }} f\left(\frac{1}{2}\right)$ is:
[JEE Main 2019]
1) $\frac{6}{25}$
2) $\frac{4}{5}$
3) $\frac{24}{25}$
4) $\frac{18}{25}$

Solution:

$
\int_0^x f(t) d t=x^2+\int_x^1 t^2 f(t) d t
$
Differentiate wrt $x$ :

$
\begin{aligned}
f(x) & =2 x+0-x^2 f(x) \\
f(x) & =\frac{2 x}{x^2+1} \\
f^{\prime}(x) & =\frac{\left(1+x^2\right) 2-(2 x)(2 x)}{\left(x^2+1\right)^2} \\
& =\frac{2 x^2-4 x^2+2}{\left(x^2+1\right)^2}
\end{aligned}
$

f^{\prime}\left(\frac{1}{2}\right)=\frac{24}{25}

Hence, the answer is the option (3).

Example 3: Let $f$ be a differentiable function such that $f(1)=2$ and $f^{\prime}(x)=f(x)_{\text {for all }}$ $x \equiv \mathbf{R}$. If $h(x)=f(f(x))$, then $h^{\prime}(1)$ is equal to: [JEE Main 2019]
1) $4 e^2$
2) $2 e$
3) $4 e$
4) $2 e^2$

Solution:
Derivative at a point-
The value of $f(x)$ obtained by putting $x=a$ is called the derivative of $f(x)$ at $x=a$ and it is denoted by $f(a)$ or
$\frac{d y}{d x}$ at $x=a$

$\begin{aligned} & \frac{d y}{d x} \text { at } x=\mathrm{a} . \\ & \frac{d t}{d x}=f \log |f|=x+c x=1 \quad t-=2 \log 2=1+c \quad c=\log 2-1 \log |f|=x+\log 2-1= \\ & \quad h^{\prime}(x)=f^{\prime}(f(x)) \times f^{\prime}(x) \\ & h^{\prime}(1)=f^{\prime}(2) \times f^{\prime}(1) \\ & f^{\prime}(1)=f(1)=2 \\ & f^{\prime}(2)=2 e^{2-1}=2 e \\ & \Rightarrow h^{\prime}(1)=2 e \times 2=4 e\end{aligned}$

Hence, the answer is the option 3.

Example 4: to: [JEE Main 2023]
1) $\frac{-2}{3}$
2) $\frac{2}{9}$
3) $\frac{-1}{3 \sqrt{3}}$
4) $\frac{2}{3 \sqrt{3}}$

Solution:

$\begin{aligned} & f(x)=-\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \\ & f^{\prime}(x)=-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \\ & f^{\prime \prime}(x)=-\sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \tan \left(\frac{x}{2} \frac{-\pi}{8}\right) \cdot \frac{1}{2}\end{aligned}$

$\begin{aligned} & f\left(\frac{7 \pi}{12}\right)=-\tan \left(\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}} \\ & f^{\prime \prime}\left(\frac{7 \pi}{12}\right)=-\frac{1}{2} \sec ^2 \frac{\pi}{6} \cdot \tan \frac{\pi}{6}=\frac{-1}{2} \cdot \frac{4}{3} \times \frac{1}{\sqrt{3}}=\frac{-2}{3 \sqrt{3}} \\ & f\left(\frac{7 \pi}{12}\right) \cdot f^{\prime \prime}\left(\frac{7 \pi}{12}\right)=\frac{2}{9}\end{aligned}$

Hence, the answer is the option (2).

Example 5: For the differentiable function $f: R-\{0\} \rightarrow R$, let $3 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-10$, $\left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|_{\text {is equal to: }}$
[JEE Main 2023]
1) 13
2) $\frac{29}{5}$
3) $\frac{33}{5}$
4) 1

Solution:

$\begin{aligned} & {\left[3 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-10\right] \times 3} \\ & {\left[2 f(x)+3 f\left(\frac{1}{x}\right)=x-10\right] \times 2} \\ & 5 f(x)=\frac{3}{x}-2 x-10 \\ & f(x)=\frac{1}{5}\left(\frac{3}{x}-2 x-10\right)\end{aligned}$

$\begin{aligned} & f^{\prime}(x)=\frac{1}{5}\left(-\frac{3}{x^2}-2\right) \\ & \left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|=\left|\frac{1}{5}(1-6-10)+\frac{1}{5}(-48-2)\right| \\ & =|-3-10|=13\end{aligned}$

Hence, the answer is the option 1

Summary

Differentiation is an important concept of Calculus. Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point. With the help of differentiation, we can find the rate of change of one quantity for another. The concept of differentiation is the cornerstone on which the development of calculus rests.