Differentiation Rules: Definition, Formula, Examples

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 of some basic functions

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

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

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

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

5. $\frac{d}{d x}\left(\log _{\mathrm{e}}|\mathbf{x}|\right)=\frac{\mathbf{1}}{\mathbf{x}}, \quad \mathbf{x} \neq 0$

6. $\quad \frac{d}{d x}\left(\log _{\mathbf{a}}|\mathbf{x}|\right)=\frac{1}{\mathbf{x} \log _{\mathrm{e}} \mathbf{a}}, \quad \mathbf{x} \neq 0$

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

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

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

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

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

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

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

Basic Differentiation Rules

The basic differentiation rules are essential tools in Class 12 calculus. They allow students to find derivatives of various functions efficiently and are widely used in board exams, JEE, CUET, and other competitive exams. Understanding these rules makes it easier to handle polynomials, trigonometric, exponential, logarithmic, and composite functions.

Sum and Difference Rule

The sum and difference rule states that the derivative of the sum or difference of two functions equals the sum or difference of their individual derivatives.

$\frac{d}{dx}[f(x)+g(x)] = f'(x)+g'(x)$

$\frac{d}{dx}[f(x)-g(x)] = f'(x)-g'(x)$

Example: If $f(x) = x^3$ and $g(x) = \cos x$, then

$\frac{d}{dx}[x^3 + \cos x] = 3x^2 - \sin x$

This rule simplifies complex expressions by allowing you to differentiate term by term.

Constant Multiple Rule

The constant multiple rule states that a constant multiplied by a function can be factored out of the derivative.

$\frac{d}{dx}[k f(x)] = k f'(x)$

Example: If $f(x) = x^2$ and $k = 4$, then

$\frac{d}{dx}[4 x^2] = 4 \cdot 2x = 8x$

This rule is very useful when dealing with scaled functions in algebra, physics, or economics problems.

Product Rule

The product rule is used to differentiate the product of two functions. It ensures that both functions’ contributions to the derivative are accounted for.

$\frac{d}{dx}[f(x) g(x)] = f(x) g'(x) + g(x) f'(x)$

Example: If $f(x) = x^2$ and $g(x) = \sin x$, then

$\frac{d}{dx}[x^2 \sin x] = x^2 \cos x + 2x \sin x$

Extension to three or more functions:
For $h(x) = f(x) g(x) k(x)$,

$h'(x) = f'(x) g(x) k(x) + f(x) g'(x) k(x) + f(x) g(x) k'(x)$

This rule is essential for polynomial-trigonometric or exponential-trigonometric combinations.

Quotient Rule

The quotient rule applies when differentiating a function divided by another function. It is key for ratios of polynomials, trigonometric functions, and exponential functions.

$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2}$

Example: If $f(x) = x^2$ and $g(x) = \sin x$, then

$\frac{d}{dx}\left(\frac{x^2}{\sin x}\right) = \frac{\sin x \cdot 2x - x^2 \cdot \cos x}{\sin^2 x}$

This rule is widely used in optimization, rate-of-change problems, and applied calculus.

Chain Rule

The chain rule is used to differentiate composite functions. It allows you to compute derivatives of functions nested inside other functions, which is common in trigonometric, exponential, and logarithmic problems.

For $y = f(g(x))$:

$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

Example: If $y = \sin(x^2)$, then

$\frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x \cos(x^2)$

Extension to multiple nested functions:
For $y = f(g(h(x)))$,

$\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$

This rule is essential for differentiating complex composite functions in board exams and competitive tests.

Power Rule

The power rule is used to differentiate functions of the form $f(x) = x^n$, where $n$ is any real number. It is one of the simplest and most frequently applied rules in calculus.

$\frac{d}{dx}[x^n] = n x^{n-1}$

Example 1: If $f(x) = x^5$, then

$\frac{d}{dx}[x^5] = 5x^4$

Example 2: If $f(x) = \frac{1}{x^3} = x^{-3}$, then

$\frac{d}{dx}[x^{-3}] = -3 x^{-4} = -\frac{3}{x^4}$

The power rule also extends to constants multiplied by powers, so for $f(x) = k x^n$,

$\frac{d}{dx}[k x^n] = k n x^{n-1}$

This rule is essential for polynomial functions, algebraic expressions, and derivative problems in board exams and competitive exams.

Differentiation of Special Functions

Some functions require specific differentiation formulas because they do not follow the basic power or sum rules directly. These include trigonometric, exponential, logarithmic, inverse trigonometric functions, as well as functions defined implicitly or parametrically.

Derivatives of Trigonometric Functions

The trigonometric functions are commonly used in calculus. Their derivatives are standard and must be memorized for quick problem-solving:

$\frac{d}{dx}[\sin x] = \cos x$

$\frac{d}{dx}[\cos x] = -\sin x$

$\frac{d}{dx}[\tan x] = \sec^2 x$

$\frac{d}{dx}[\cot x] = -\csc^2 x$

$\frac{d}{dx}[\sec x] = \sec x \tan x$

$\frac{d}{dx}[\csc x] = -\csc x \cot x$

Example: If $y = \sin x + \tan x$, then

$\frac{dy}{dx} = \cos x + \sec^2 x$

Derivatives of Exponential and Logarithmic Functions

Exponential and logarithmic functions are widely used in growth, decay, and financial mathematics.

$\frac{d}{dx}[e^x] = e^x$

$\frac{d}{dx}[a^x] = a^x \ln a$

$\frac{d}{dx}[\ln x] = \frac{1}{x}$

$\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}$

Example: If $y = 3^x + \ln x$, then

$\frac{dy}{dx} = 3^x \ln 3 + \frac{1}{x}$

Derivatives of Inverse Trigonometric Functions

The inverse trigonometric functions often appear in integration and geometry problems.

$\frac{d}{dx}[\sin^{-1} x] = \frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}[\cos^{-1} x] = -\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}[\tan^{-1} x] = \frac{1}{1+x^2}$

$\frac{d}{dx}[\cot^{-1} x] = -\frac{1}{1+x^2}$

Example: If $y = \tan^{-1} x + \sin^{-1} x$, then

$\frac{dy}{dx} = \frac{1}{1+x^2} + \frac{1}{\sqrt{1-x^2}}$

Implicit Differentiation

Sometimes, $y$ is not given explicitly in terms of $x$. In such cases, implicit differentiation is used.

Example: If $x^2 + y^2 = 25$, differentiate both sides w.r.t $x$:

$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$

This technique is crucial for circles, ellipses, and other curves.

Parametric Differentiation

When functions are defined in parametric form, derivatives are calculated using $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

Example: If $x = t^2 + 1$ and $y = t^3$, then

$\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 3t^2$

$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$

Parametric differentiation is essential for motion problems, curve analysis, and applied mathematics.

Important Formulae for Differentiation

This section lists all essential differentiation formulas that Class 12 students must know for board exams, JEE, CUET, and other competitive exams. From power, trigonometric, exponential, logarithmic, inverse functions to composite, product, and quotient rules, these formulas help solve problems quickly and accurately.

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.

List of topics related to differentiation rules

This section covers all important differentiation topics for Class 12, helping students master key concepts and prepare effectively for board exams and competitive tests.

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

This section lists reliable NCERT resources including notes, solutions, and exemplar problems to help Class 12 students strengthen differentiation rules concepts.

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 Rules

Test your knowledge with carefully designed practice questions and MCQs, improving speed, accuracy, and problem-solving for board exams, JEE and CUET.

Differentiation Rules- Practice Question MCQ

We have shared below the links to practice questions on related topics of differentiaion rules: