Differentiation: Definition, Rule, Formula, Examples

How do epidemiologists figure out the increase or decrease of an infection? It is all with the help of differentiation. Differentiation is a concept of rate of change. Differentiation is used to calculate the rate of change of one quantity with respect to another quantity.

Differentiation and integration are 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. What is Differentiation?
2. Differentiation Formulas
3. Rules of Differentiation
4. Solved Examples Based on Rules of Differentiation

This article is about the concept of differentiation class 11. This concept falls under the broader category of Calculus. 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 nine questions have been asked on this concept, including one in 2013, one in 2018, two in 2019, three in 2020, one in 2021, and one in 2022.

What is 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 $\lim _{\Delta x \rightarrow 0} \frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}$ 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)=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}
$

For all $x$ for which this limit exists,
$f^{\prime}(x)=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$ 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 $f^{\prime}(x), \frac{d y}{d x}, y^{\prime}, \frac{d}{d x}[f(x)], D_x[y]$ or $D y$ or $y_1$. Here $\frac{d}{d x}$ or $D$ is the differential operator.

Differentiation Formulas

Differentiation formulas include differentiation of some basic functions and the rules of differentiation.

Differentiation of Some Basic Functions

Differentiation of some basic functions includes differentiation of a constant, basic polynomials, logarithms and trigonometric functions.

Differentiation of a Constant

• The differentiation of a constant $a$ is $\frac{d}{d x}(a)=0$

Differentiation of Basic Polynomials

The differentiation of some basic polynomials are

• The differentiation of $x^n$ is $nx^{n-1}$
• The differentiation of $a^x$ is $a^x \log_e a$

Differentiation of Logarithms

The differentiation of logarithms are

• The differntiation of $e^x$ is $\frac{d}{d x}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{e}^{\mathrm{x}} \log _{\mathrm{e}} \mathrm{e}=\mathrm{e}^{\mathrm{x}}$
• The differentiation of log x is $\frac{d}{d x}\left(\log\mathbf{x}\right)=\frac{\mathbf{1}}{\mathbf{x}}, \quad \mathbf{x} \neq 0$
• The differentiation of $\log _{\mathrm{a}}|x|$ is $\frac{d}{d x}\left(\log _{\mathbf{a}}|\mathbf{x}|\right)=\frac{1}{\mathbf{x} \log _{\mathrm{e}} \mathbf{a}}, \quad \mathbf{x} \neq 0$

Differentiation of Trigonometric Functions

The differentiation of trigonometric functions are

• The differentiation of sin x is $\frac{d}{d x}(\sin (\mathbf{x}))=\cos (\mathbf{x})$
• The differentiation of cos x is $\frac{d}{d x}(\cos (\mathbf{x}))=-\sin (\mathbf{x})$
• The differentiation of tan x is $\frac{d}{d x}(\tan (\mathbf{x}))=\sec ^2(\mathbf{x})$
• The differentiation of cosec x is $\frac{d}{d x}(\csc (\mathbf{x}))=-\csc (\mathbf{x}) \cot (\mathbf{x})$
• The differentiation of sec x is $\frac{d}{d x}(\sec (\mathbf{x}))=\sec (\mathbf{x}) \tan (\mathbf{x})$
• The differentiation of cot x is $\frac{d}{d x}(\cot (\mathbf{x}))=-\csc ^2(\mathbf{x})$

Differentiation of Inverse Trigonometric Functions

• The differentiation of $sin^{-1} x$ is $\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}}, x \neq \pm 1$
• The differentiation of $cos^{-1} x$ is $\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^2}}, x \neq \pm 1$
• The differentiation of $tan^{-1} x$ is $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$
• The differentiation of $cosec^{-1} x$ is $\frac{d}{d x}\left(\csc ^{-1} x\right)=\frac{-1}{|x| \sqrt{x^2-1}}, x \neq \pm 1$
• The differentiation of $sec^{-1} x$ is $\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{|x| \sqrt{x^2-1}}, x \neq \pm 1$
• The differentiation of $cot^{-1} x$ is $\frac{d}{d x}\left(\cot ^{-1} x\right)-\frac{-1}{1+x^2}$

Rules of Differentiation

The important rules of differentiation are

• Power Rule
• Sum and Difference Rule
• Product Rule
• Quotient Rule
• Chain Rule

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))
$
In general,

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

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 $g$.

$
\begin{aligned}
& \frac{d}{d x}(f(x)-g(x))=\frac{d}{d x}(f(x))-\frac{d}{d x}(g(x)) \\
& \frac{d}{d x}(f(x)-g(x)-h(x)-\ldots \ldots)=\frac{d}{d x}(f(x))-\frac{d}{d x}(g(x))-\frac{d}{d x}(h(x))-\ldots \ldots
\end{aligned}
$

Constant Multiple Rule

The derivative of a constant $k$ multiplied by a function $f$ is the same as the constant multiplied by the derivative of $f$

$
\frac{d}{d x}(k f(x))=k \frac{d}{d x}(f(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))
$

This means that the product differentiation of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Extending the Product Rule

If three functions are involved, i.e let $k(x)=f(x) \cdot g(x) \cdot h(x)$
Let us have a function $\mathrm{k}(\mathrm{x})$ as the product of the function $\mathrm{f}(\mathrm{x}), \mathrm{g}(\mathrm{x})$ and $\mathrm{h}(\mathrm{x})$. That is, $k(x)=(f(x) \cdot g(x)) \cdot h(x)$. Thus,

$
k^{\prime}(x)=\frac{d}{d x}(f(x) g(x)) \cdot h(x)+\frac{d}{d x}(h(x)) \cdot(f(x) g(x))
$

[By applying the product rule to the product of $f(x) g(x)$ and $h(x)$.]

$
\begin{aligned}
& =\left(f^{\prime}(x) g(x)+g^{\prime}(x) f(x)\right) h(x)+h^{\prime}(x) f(x) g(x) \\
& =f^{\prime}(x) g(x) h(x)+f(x) g^{\prime}(x) h(x)+f(x) g(x) h^{\prime}(x)
\end{aligned}
$

Quotient Rule

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

$
\frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\frac{g(x) \cdot \frac{d}{d x}(f(x))-f(x) \cdot \frac{d}{d x}(g(x))}{(g(x))^2}
$
OR
if $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime}(x)=\frac{f^{\prime}(x) g(x)-g^{\prime}(x) f(x)}{(g(x))^2}$
As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives.

Chain Rule

If $u(x)$ and $v(x)$ are differentiable functions, then $u o v(x)$ or $u[v(x)] {\text { is also differentiable. }}$
If $y=u o v(x)=u[v(x)]$, then

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} u\{v(x)\}}{\mathrm{d}\{v(x)\}} \times \frac{\mathrm{d}}{\mathrm{d} x} v(x)
$

is known as the chain rule. Or,

$
\text { If } y=f(u) \text { and } u=g(x) \text {, then } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} u} \cdot \frac{\mathrm{d} u}{\mathrm{~d} x}
$
The chain rule can be extended as follows
If $y=[\operatorname{uovow}(x)]=u[v\{w(x)\}]$, then

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}[u[v\{w(x)\}]}{\mathrm{d} v\{w(x)\}} \times \frac{\mathrm{d}[v\{w(x)\}]}{\mathrm{d} w(x)} \times \frac{\mathrm{d}[w(x)]}{\mathrm{d} x}
$

Recommended Video Based on Rules of Differentiation

Solved Examples Based on Rules of Differentiation

Example 1: Let $y=\sin ^2 x+2 \cos ^3 2 x$, then $\mathrm{dy} / \mathrm{dx}$ equals
1) $
\sin 2 x+12 \sin 2 x \cos ^2 2 x
$

2) $
\cos ^{2 x}+6 \cos ^2 2 x
$

3) $
\sin 2 x-12 \sin 2 x \cos ^2 2 x
$

4) $
\cos ^{2 x}+6 \sin ^2 2 x
$

Solution:

The rule for differentiation-
The derivative of the sum or difference of two functions is the sum or difference of their derivatives.

$
\begin{aligned}
& \frac{d}{d x} f(x) \pm g(x)=\frac{d}{d x} f(x) \pm \frac{d}{d x} g(x) \\
& y=\sin ^2 x+2 \cos ^3 2 x \Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(\sin ^2 x+2 \cos ^3 2 x\right) \\
& =\frac{d}{d x} \sin ^2 x+\frac{d}{d x} 2 \cos ^3 2 x=\frac{d}{d x}(\sin x)^2+2 \frac{d}{d x}(\cos 2 x)^3 \\
& =\frac{d(\sin x)}{d x} \frac{d\left(\sin ^2 x\right)}{d(\sin x)}+2 \frac{d(\cos 2 x)^3}{d(\cos 2 x)} \cdot \frac{d(2 x)}{d x} \\
& =2 \sin x \cos x+2 * 3(\cos x)^2 *(-\sin 2 x) * 2 \\
& =\sin 2 x-12 \sin 2 x \cos ^2 x
\end{aligned}
$

Hence, the answer is the option 3.

Example 2: The value of $\log _e 2 \frac{d}{d x}\left(\log _{\cos x} \operatorname{cosec} x\right)$ at $x=\frac{\pi}{4}$ is
[JEE Main 2022]
1) $-2 \sqrt{2}$
2) $2 \sqrt{2}$
3) $-4$
4) $4$

Solution:

$
\begin{aligned}
& \log 2 \cdot \frac{d}{d x}\left(\frac{\log (\operatorname{cosec} x)}{\log (\cos x)}\right) \\
& =-\log 2 \cdot \frac{d}{d x}\left(\frac{\log (\sin x)}{\log (\cos x)}\right) \\
& =-\log 2 \cdot \frac{\log (\cos x) \cdot \frac{\cos x}{\sin x}-\log (\sin x) \frac{(-\sin x)}{\cos x}}{(\log \cos x)^2}
\end{aligned}
$
At $\mathrm{x}=\frac{\pi}{4}$

$
\begin{aligned}
& =-\log 2 \cdot \frac{\log \left(\frac{1}{\sqrt{2}}\right)+\log \left(\frac{1}{\sqrt{2}}\right)}{\left(\log \left(\frac{1}{\sqrt{2}}\right)\right)^2} \\
& =-\log 2 \cdot \frac{\left(2 \log \left(\frac{1}{\sqrt{2}}\right)\right)}{\left(\log \left(\frac{1}{\sqrt{2}}\right)\right)^2}
\end{aligned}
$
$\begin{aligned} & =\frac{-\log 2 \cdot 2}{\log \left(\frac{1}{\sqrt{2}}\right)} \\ & =\frac{-2 \log 2}{-\log (\sqrt{2})} \\ & =\frac{2 \log 2}{\frac{1}{2} \log 2} \\ & =4\end{aligned}$
Hence, the answer is the option (4).

Example 3: The minimum value of $\alpha$ for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha$ has at least one solution in $\left(0, \frac{\pi}{2}\right)$ is
[JEE Main 2021]
1) $7$
2) $8$
3) $9$
4) $10$

Solution:
Let $f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$

$
y=\frac{4-3 \sin x}{\sin x(1-\sin x)}
$
Let $\sin \mathrm{x}=\mathrm{t}$ when $t \in(0,1)$.

$
y=\frac{4-3 t}{t-t^2}
$

$\begin{aligned} & \frac{d y}{d t}=\frac{-3\left(t-t^2\right)-(1-2 t)(4-3 t)}{\left(t-t^2\right)^2}=0 \\ & \Rightarrow 3 \mathrm{t}^2-3 \mathrm{t}-\left(4-11 \mathrm{t}+6 \mathrm{t}^2\right)=0 \\ & \Rightarrow 3 \mathrm{t}^2-8 \mathrm{t}+4=0 \\ & \Rightarrow 3 \mathrm{t}^2-6 \mathrm{t}-2 \mathrm{t}+4=0 \\ & \Rightarrow \mathrm{t}=\frac{2}{3} \quad \text { and } \quad \mathrm{t} \neq 2 \\ & \frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha \\ & \frac{12}{2}+\frac{3}{3-2}=\alpha \\ & \alpha=9\end{aligned}$

Hence, the answer is the option 3.

Example 4: Let $f$ and $g$ be differentiable functions on $R$ such that $f o g$ is the identity function. If for some $a, b \in \mathbf{R}, g^{\prime}(a)=5$ and $g(a)=b {\text { then }} f^{\prime}(b)$ is equal to [JEE Main 2020]
1) $\frac{2}{5}$
2) $5$
3) $1$
4) $\frac{1}{5}$

Solution:

Let $f$ and $g$ be functions. For all $x$ in the domain of $g$ for which $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, the derivative of the composite function

$
\begin{aligned}
& h(x)=(f \circ g)(x)=f(g(x)) \text { Is given by } \\
& h^{\prime}(x)=f^{\prime}(g(x)) \cdot g^{\prime}(x)
\end{aligned}
$
Composites of Three or More Functions
For all values of $x$ for which the function is differentiable, if $k(x)=h(f(g(x)))$ Then,

$
\begin{aligned}
& k^{\prime}(x)=h^{\prime}(f(g(x))) \cdot f^{\prime}(g(x)) \cdot g^{\prime}(x) \\
& f(g(x))=x \\
& \Rightarrow f^{\prime}(g(x)) \cdot g^{\prime}(x)=1 \\
& P u t x=a \\
& \Rightarrow f^{\prime}(g(a)) g^{\prime}(a)=1 \\
& \Rightarrow f^{\prime}(b) \times 5=1 \Rightarrow f^{\prime}(b)=\frac{1}{5}
\end{aligned}
$

Example 5: If $f(x)=\sin ^{-1}\left(\frac{2 \times 3^x}{1+9^x}\right)$, then $f^{\prime}\left(-\frac{1}{2}\right)$ equals
[JEE Main 2018]
1) $-\sqrt{3} \log _e \sqrt{3}$
2) $\sqrt{3} \log \sqrt{3}$
3) $-\sqrt{3} \log _e 3$
4) $\sqrt{3} \log _e 3$

Solution:
As we learned,
Chain Rule for differentiation (indirect) -
Let $y=f(x)$ is not in standard form then

$
\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}
$

Now

$\begin{aligned} & f(x)=\sin ^{-1}\left(\frac{2 \times 3^x}{1+9^x}\right) \\ & =2 \tan ^{-1} 3^x \\ & \because 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^2} \quad \text { if }-1 \leq x \leq 1 \\ & \frac{d}{d u}(\arctan (u))=\frac{1}{u^2+1} \\ & \frac{d}{d x}\left(3^x\right)=\ln (3) \cdot 3^x \\ & f^{\prime}(x)=2 \times \frac{1}{1+\left(3^x\right)^2} \times 3^x \times \ln 3 \\ & f^{\prime}\left(\frac{-1}{2}\right)=2 \times \frac{1}{1+\left(3^{-1}\right)} \times 3^{\frac{-1}{2}} \times \ln 3 \\ & =2 \times \frac{3}{4} \times \frac{1}{\sqrt{3}} \times \ln 3 \\ & =\sqrt{3} \times \frac{1}{2} \ln 3\end{aligned}$

Hence, the answer is the option 2.