Differentiation: Definition, Rule, Formula, Examples

What is Differentiation?

The process of finding the derivative of a function is called differentiation. Let $f$ be defined on an open interval $I$ containing a point $x_0$, and suppose that the limit

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

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

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

For all $x$ for which this limit exists, the derivative is written as

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

The derivative represents the instantaneous rate of change of the function and is a key concept in Class 12 differentiation.

The derivative of $y = f(x)$ can also be written using different notations:

$f'(x),\ \frac{dy}{dx},\ y',\ \frac{d}{dx}[f(x)],\ D_x[y],\ Dy,\ y_1$

Here, $\frac{d}{dx}$ or $D$ is called the differential operator.

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 let $k$ be a constant. The following differentiation rules hold for Class 12 Calculus and are widely used in JEE, and CUET problems.

Sum Rule

The derivative of the sum of two functions is the sum of their derivatives:

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

In general, for multiple functions:

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

Difference Rule

The derivative of the difference of two functions is the difference of their derivatives:

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

For multiple functions:

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

Constant Multiple Rule

The derivative of a constant multiplied by a function is the constant times the derivative of the function:

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

Product Rule

For differentiable functions $f(x)$ and $g(x)$:

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

If three functions are multiplied, $k(x) = f(x) g(x) h(x)$, the extended product rule is:

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

Quotient Rule

For differentiable functions $f(x)$ and $g(x)$:

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

Or, if $h(x) = \frac{f(x)}{g(x)}$, then

$h'(x) = \frac{f'(x) g(x) - g'(x) f(x)}{(g(x))^2}$

Note: The derivative of a quotient is not the quotient of the derivatives.

Chain Rule

If $u(x)$ and $v(x)$ are differentiable functions, the composition $y = u(v(x))$ is differentiable, and:

$\frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}$

Or equivalently, if $y = f(u)$ and $u = g(x)$:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

For three nested functions $y = u(v(w(x)))$:

$\frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}$

Differentiation of Special Functions

Some functions in calculus are not written in simple form like $y = f(x)$. For such cases, we use special differentiation techniques. These include implicit differentiation, parametric differentiation, and differentiation using product and quotient rules. These concepts are frequently asked in Class 12 board exams and competitive exams like JEE and CUET.

Differentiation of Implicit Functions

An implicit function is one in which $x$ and $y$ are mixed together in the same equation, such as $x^2 + y^2 = 25$. To differentiate such functions, we differentiate both sides with respect to $x$ and treat $y$ as a function of $x$.

Example of differentiation of implicit functions:
$2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$.

Differentiation of Parametric Functions

In parametric differentiation, both $x$ and $y$ are expressed in terms of a third variable, usually $t$. Instead of direct differentiation, we use the formula
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
For example, if $x = \cos t$ and $y = \sin t$, then $\frac{dy}{dx} = -\cot t$.

Important Differentiation Formulae List

This section provides a comprehensive list of key differentiation formulas for Class 12, covering standard functions, trigonometric functions, exponential, logarithmic, and higher-order derivatives for quick reference and exam preparation.

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

This section outlines all important differentiation topics for Class 12, helping students focus on key concepts for board exams and competitive exams.

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

Use these NCERT notes, solutions, and exemplar problems to strengthen your understanding of differentiation 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

Test your understanding of Class 12 differentiation concepts with carefully designed MCQs. These practice questions help improve speed, accuracy, and problem-solving skills for board exams and competitive tests like JEE and CUET.

Differentiation- Practice Question MCQ

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