Differentiability and Existence of Derivative: Difference, Examples

What is a Derivative?

The derivative of a function represents the instantaneous rate of change of a function with respect to its independent variable. In simple terms, it measures how fast a function’s output changes when the input changes by a small amount.

Let $f(x)$ be a function of one variable and $\Delta x$ be a small increment (positive or negative) in $x$. The corresponding change in the function is given by

$\Delta f = f(x + \Delta x) - f(x)$

If $\frac{\Delta f}{\Delta x}$ approaches a finite limit as $\Delta x \rightarrow 0$, then this limit is called the derivative of $f$ at $x$.

It is denoted as $f'(x)$, $\frac{df}{dx}$, or $\frac{d}{dx}f(x)$. Hence,

$\frac{df}{dx} = \lim\limits_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$

When evaluated at a specific point $x = a$, the derivative is written as $f'(a)$ or $\left(\frac{df}{dx}\right)_{x=a}$.

Graphical Representation of a Derivative

Let $P(x, y)$ be any point on the curve $y = f(x)$ and $Q(x + \delta x, y + \delta y)$ be another point close to $P$.

Then, the slope of the tangent to the curve at $(x, y)$ is given by

$M_T = \lim\limits_{\delta x \rightarrow 0} \frac{(y + \delta y) - y}{(x + \delta x) - x} = \lim\limits_{\delta x \rightarrow 0} \frac{\delta y}{\delta x}$

Therefore,

$M_T = \left(\frac{dy}{dx}\right)$ at $(x, y)$

The process of finding the derivative of a function is called differentiation.

Condition for Differentiability

While defining a derivative, we always assume that the limit must exist. But what happens when it doesn’t?

A function $f(x)$ is said to be differentiable at a point $x = c$ if the limit

$\lim\limits_{h \rightarrow 0} \frac{f(c + h) - f(c)}{h}$

exists and is finite.

If this limit does not exist, the function is not differentiable at $x = c$.

In other words, a function $f(x)$ is differentiable at a point $x = a$ if both the following limits are finite and equal:

$\lim\limits_{h \rightarrow 0^{+}} \frac{f(a + h) - f(a)}{h}$ and $\lim\limits_{h \rightarrow 0^{-}} \frac{f(a + h) - f(a)}{h}$

Right-Hand and Left-Hand Derivatives

The right-hand derivative (RHD) and left-hand derivative (LHD) of $f(x)$ at $x = a$ are defined as:

RHD: $R f'(a) = \lim\limits_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}$

LHD: $L f'(a) = \lim\limits_{h \rightarrow 0} \frac{f(a - h) - f(a)}{-h}$

The function $f(x)$ is differentiable at $x = a$ if both $R f'(a)$ and $L f'(a)$ are finite and equal, that is,

$R f'(a) = L f'(a)$

Hence,

$\lim\limits_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h} = \lim\limits_{h \rightarrow 0} \frac{f(a - h) - f(a)}{-h}$

Geometrical Meaning of Derivative

This section explains the geometrical meaning of a derivative, showing how the derivative represents the slope of the tangent to a curve at a given point. It covers the concepts of right-hand and left-hand derivatives, helping you visualize how differentiability is linked to the smoothness and direction of a curve.

(a) Right-Hand Derivative

Let $A(a, f(a))$ and $B(a + h, f(a + h))$ be two points on the curve $y = f(x)$.

The slope of the chord AB is given by

$\text{Slope of } AB = \frac{f(a + h) - f(a)}{(a + h) - a}$

Taking limit as $h \rightarrow 0$, we get

$\lim\limits_{h \rightarrow 0} (\text{Slope of } AB) = \lim\limits_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}$

Thus,

$R[f'(a)] = \lim\limits_{h \rightarrow 0} (\text{Slope of chord } AB)$

As $h \rightarrow 0$, point $B$ approaches $A$, and the chord becomes the tangent drawn at $A$ towards the right.

Hence, the right-hand derivative represents the slope of the tangent to the curve at $A$ towards the right.

(b) Left-Hand Derivative

Let $A(a, f(a))$ and $C(a - h, f(a - h))$ be two nearby points on the curve $y = f(x)$.

The slope of chord AC is

$\text{Slope of } AC = \frac{f(a - h) - f(a)}{(a - h) - a}$

Taking limit as $h \rightarrow 0$, we get

$\lim\limits_{h \rightarrow 0} (\text{Slope of } AC) = \lim\limits_{h \rightarrow 0} \frac{f(a - h) - f(a)}{-h}$

Thus,

$L[f'(a)] = \lim\limits_{h \rightarrow 0} (\text{Slope of chord } AC)$

As $h \rightarrow 0$, point $C$ approaches $A$, and the chord becomes the tangent drawn at $A$ towards the left.

Hence, the left-hand derivative represents the slope of the tangent to the curve at $A$ towards the left.

Relationship Between Continuity and Differentiability

Understanding the connection between continuity and differentiability is one of the most important concepts in Class 12 Mathematics Chapter 5 – Continuity and Differentiability. Both ideas are closely linked, but they are not the same. Differentiability is a stronger condition than continuity, meaning every differentiable function is continuous, but the reverse is not always true.

Every Differentiable Function is Continuous

If a function $f(x)$ is differentiable at a point $x = a$, then it must also be continuous at that point. In simple terms, if the derivative $f'(a)$ exists, there can’t be any breaks or jumps in the graph of $f(x)$ near $x = a$.

Mathematically, if $\lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}$ exists and is finite,
then $\lim\limits_{x \to a} f(x) = f(a)$,
which ensures continuity at $x = a$.

This relationship highlights that differentiability implies continuity, but continuity alone does not guarantee differentiability.

Geometrical Meaning of the Existence of Derivative

For a function $f(x)$ to have a derivative at $x = a$, both the left-hand and right-hand tangents at that point must coincide.

That is, $L[f'(a)] = R[f'(a)]$.

This means:

1. The slope of the tangent drawn at $A$ from the left equals that from the right.

2. There exists a unique tangent at $x = a$.

3. The curve is smooth and continuous around the point $x = a$.

Solved Examples Based On Tangents to the curve at a point:

Example 1: Let $S$ be the set of all points in $(-\pi, \pi)$ at which the function, $f(x)=\min \{\sin x, \cos x\}$ is not differentiable. Then S is a subset of which of the following?
[JEE Main 2019]
1) $\left\{-\frac{3 \pi}{4},-\frac{\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{4}\right\}$
2) $\left\{-\frac{3 \pi}{4},-\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{\pi}{4}\right\}$
3) $\left\{-\frac{\pi}{4}, 0, \frac{\pi}{4}\right\}$
4) $\left\{-\frac{\pi}{2},-\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{2}\right\}$

Solution

Differentiability -
Let $\mathrm{f}(\mathrm{x})$ be a real valued function defined on an open interval ( $\mathrm{a}, \mathrm{b})$ and $x \in(\mathrm{a}$, b). Then the function $\mathrm{f}(\mathrm{x})$ is said to be differentiable at $x_{\circ}$ if

$
\lim\limits_{h \rightarrow 0} \frac{f\left(x_0+h\right)-f\left(x_0\right)}{\left(x_0+h\right)-x_0}
$

or $\lim\limits_{x \rightarrow x_0} \frac{f(x)-f\left(x_0\right)}{x-x_0}$
-Hence number of points $f(x)$ is non-differentiable are 2 which are $\frac{-3 \pi}{4}$ and $\frac{\pi}{4}$

Example 2: Let $f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^2\right\}, & |x| \leq 2 \\ 8-2|x| & 2<|x| \leq 4 \text { Let } \mathrm{S} \text { be the }\end{array}\right.$ set of points in the interval $(-4,4)$ at which f is not differentiable. Then S : [JEE Main 2019]
1) is an empty set
2) equals $\{-2,-1,0,1,2\}$
3) equals $\{-2,-1,1,2\}$
4) equals $\{-2,2\}$

Solution

Condition for differentiability -
A function $\mathrm{f}(\mathrm{x})$ is said to be differentiable at $x=x_{\circ}$ if $R f^{\prime}\left(x_{\circ}\right)$ and $L f^{\prime}\left(x_{\circ}\right)$ both exist and are equal otherwise non differentiable

Geometrical interpretation of Derivative -
Let $P$ be any point $(x, y)$ on the curve $y=f(x)$ and $Q$ is a point in the neighbourhood of $P$ on either side of $P$. such that the co-ordinate of the point Q are
$(x+\delta x, y+\delta y)$ satisfying the curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$
$\therefore M_T=$ slope of tangent

$
\begin{aligned}
& =\lim\limits_{\delta x \rightarrow 0} \frac{(y+\delta y)-y}{(x+\delta x)-x}=\lim\limits_{\delta x \rightarrow 0} \frac{\delta y}{\delta x} \\
& \therefore \quad M_T=\left(\frac{d y}{d x}\right) \text { at }(x, y)
\end{aligned}
$

- wherein

Where ( $x, y$ ) on the curve and $M_T$ is tangent at ( $x, y$ ).
$
f(x)=\left\{\begin{array}{ccc}
8+2 x, & & -4 \leq x<-2 \\
x^2, & & -2 \leq x \leq-1 \\
|x|, & & -1<x<1 \\
x^2, & & 1 \leq x \leq 2 \\
8-2 x, & & 2<x \leq 4
\end{array}\right.
$

$f(x)$ is not differentiable at $x=-2,-1,0,1,2$.

Example 3: Let $
S=\left\{(\lambda, \mu) \epsilon \mathbf{R} \times \mathbf{R}: f(t)=\left(|\lambda| e^{|t|}-\mu\right) \cdot \sin (2|t|), t \in \mathbf{R},\right\}
$

is a differentiable function. Then S is a subset of: [JEE Main 2018]
1) $\mathbf{R} \times[0, \infty]$
2) $[0, \infty] \times \mathbf{R}$
3) $\mathbf{R} \times[-\infty, 0]$
4) $[-\infty, 0] \times \mathbf{R}$

Solution

As we learned
-Condition for differentiable -
A function $\mathrm{f}(\mathrm{x})$ is said to be differentiable at $x=x_{\circ}$ if $R f^{\prime}\left(x_{\circ}\right)$ and $L f^{\prime}\left(x_{\circ}\right)$ both exist and are equal otherwise nondifferentiable

$
\begin{aligned}
& f(t)=\left(|\lambda| e^{|t|}-\mu\right) \sin (2|t|) \\
& =\left(|\lambda| e^t-\mu\right) \sin 2 t \quad t>0 \\
& =\left(|\lambda| e^{-t}-\mu\right) \sin 2 t \quad t<0
\end{aligned}
$

$f(t)$ is differentiable at $t=0$

$
\mathrm{LHD}=\mathrm{RHD}
$
$
\begin{aligned}
& |\lambda| \sin 2(0)+\left(|\lambda| e^{\circ}-\mu\right) 2 \cos 0=|\lambda| \sin 2(0)-2 \cos (0)\left(\lambda e^{-0}-\mu\right) \\
& \Rightarrow 4(|\lambda|-\mu)=0 \Rightarrow|\lambda|=\mu \\
& S \equiv(\lambda, \mu)=\{\lambda \equiv R, \mu \equiv(0, \infty)\}
\end{aligned}
$

S is a subset of $R \times(0, \infty)$
Hence, the answer is the option 1 .

Example 4: Let K be the set of all real values of x where the function $f(x)=\sin |x|-|x|+2(x-\pi) \cos |x|$ is not differentiable. Then the set K is equal to: [JEE Main 2019]
1) $\{0\}$
2) $\phi$ (an empty set)
3) $\{\pi\}$
4) $\{0, \pi\}$

Solution

Condition for differentiability -
A function $\mathrm{f}(\mathrm{x})$ is said to be differentiable at $x=x_{\circ}$ if $R f^{\prime}\left(x_{\circ}\right)$ and $L f^{\prime}\left(x_{\circ}\right)$ both exist and are equal otherwise non differentiable

Differentiability -
Let $\mathrm{f}(\mathrm{x})$ be a real valued function defined on an open interval $(\mathrm{a}, \mathrm{b})$ and $x \epsilon(\mathrm{a}$, b). Then the function $\mathrm{f}(\mathrm{x})$ is said to be differentiable at $x_{\circ}$ if

$
\begin{aligned}
& \lim\limits_{h \rightarrow 0} \frac{f\left(x_0+h\right)-f\left(x_0\right)}{\left(x_0+h\right)-x_0} \\
& \text { or } \lim\limits_{x \rightarrow x_0} \frac{f(x)-f\left(x_0\right)}{x-x_0}
\end{aligned}
$
Checking differentiability at $x=0$
for $x>0$,

$
\begin{aligned}
& f(x)=\sin x-x+2(x-\pi) \cos x \\
& f^{\prime}(x)=\cos x-1+2 \cos x-2(x-\pi) \sin x \\
& \mathrm{RHD}=f^{\prime}(0+)=1-1+2-2(-\pi) \cdot 0=2
\end{aligned}
$

for $x<0$,

$
\begin{aligned}
& f(x)=-\sin x+x+2(x-\pi) \cos x \\
& f^{\prime}(x)=-\cos x+1+2 \cos x-2(x-\pi) \sin x \\
& \text { LHD }=f^{\prime}(0-)=-1+1+2-2(-\pi) \cdot 0=2 \\
& \because \text { LHD }=\text { RHD }
\end{aligned}
$

differentiable at $x=0 \Rightarrow>$ differentiable everywhere
Hence, the answer is the option 2.

Example 5: Suppose differential function $f(x)$ satisfies the identity $f(x+y)=f(x)+f(y)+x y^2+x^2 y$, for all real x and y . If $\lim\limits_{x \rightarrow 0} \frac{(x)}{x}=1$ then $\mathrm{f}(3)$ is equal to: [JEE Main 2020]
1) 8
2) 10
3) 12
4) 14

Solution

Since, $\lim\limits_{x \rightarrow 0} \frac{f(x)}{x}$ exist $\Rightarrow f(0)=0$
Now, $f^{\prime}(x)=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$
\begin{aligned}
& =\lim\limits_{h \rightarrow 0} \frac{f(h)+x h^2+x^2 h}{h}(\text { take } y=h) \\
& =\lim\limits_{h \rightarrow 0} \frac{f(h)}{h}+\lim\limits_{h \rightarrow 0}(x h)+x^2 \\
& \Rightarrow f^{\prime}(x)=1+0+x^2 \\
& \Rightarrow f^{\prime}(3)=10
\end{aligned}
$
Hence, the answer is the option 2.

List of topics related to Differentiability and Existence of Derivative

This section outlines all key subtopics under differentiability and the existence of derivatives, helping you understand where and how a function can be smooth, continuous, or have sharp turns.

Continuity and Differentiability

Derivative of a function in Parametric Form

Derivative as Rate Measure: Definition, Formula, Examples

Differentiability of Composite Function

NCERT Resources

This section compiles all NCERT-based study materials for Chapter 5 – Continuity and Differentiability, including notes, solved questions, and exemplar problems aligned with CBSE and competitive exam needs.

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 Differentiability and Existence of Derivative

This section includes topic-wise multiple-choice and concept-based questions designed to test your understanding of differentiability and continuity in various contexts.

Differentiability And Existence Of Derivative- Practice Question MCQ

We have shared below the links to practice questions on the related topics to the Differentiability and Existence of Derivative: