How do I check differentiability at a boundary of a piecewise function?

Compute LHD from the left-piece formula and RHD from the right-piece formula. First confirm continuity at the point. Then confirm LHD equals RHD and both are finite.

Differentiability Of A Function

Q: What is the difference between LHD, RHD, and f'(a)?

LHD and RHD are one-sided derivatives - limits of the difference quotient as x approaches a from the left or right only. The two-sided derivative exists precisely when LHD equals RHD and both are finite.

Q: Does differentiability imply continuity?

Yes. If the derivative at a exists finitely, then f is continuous at a. The difference f(a+h) minus f(a) equals h times the difference quotient, and as h tends to zero, the right side tends to zero times the derivative, which is zero.

Q: Does continuity imply differentiability?

No. The absolute value function is continuous at zero but has left and right derivatives that disagree. The cube root is continuous at zero but has a vertical tangent. Some functions are continuous everywhere yet differentiable nowhere.

Q: Can a function have a derivative at exactly one point?

Yes. Construct f equal to x squared on rationals and minus x squared on irrationals; both give zero at zero and both give quotients tending to zero, so the derivative at zero exists and equals zero. Elsewhere, the function is not even continuous.

MathsLimits, Continuity And DifferentiabilityFor JEE aspirants

A function $f (x)$ is differentiable at $x = a$ if the limit $f^{'} (a) = lim_{h \to 0} \frac{f ( a + h ) - f ( a )}{h}$ exists finitely. This means the left-hand derivative (LHD) and right-hand derivative (RHD) must both exist, be finite, and be equal. Geometrically, differentiability means the graph has a well-defined non-vertical tangent at that point. Differentiability is a stronger condition than continuity: every differentiable function is continuous, but not every continuous function is differentiable (think of corners, cusps, and vertical tangents). This concept underlies every application of calculus in JEE, from tangent lines to optimisation.

Key Definitions - Quick Reference

Derivative from first principle: $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ , provided the limit is finite.
Left-hand derivative (LHD): $f_{-}^{'} (a) = lim_{h \to 0^{+}} \frac{f ( a - h ) - f ( a )}{- h} = lim_{h \to 0^{-}} \frac{f ( a + h ) - f ( a )}{h}$ .
Right-hand derivative (RHD): $f_{+}^{'} (a) = lim_{h \to 0^{+}} \frac{f ( a + h ) - f ( a )}{h}$ .
Differentiable at $a$ : LHD $=$ RHD $=$ finite value.
Differentiable on $[a, b]$ : differentiable at every point of $(a, b)$ , with RHD at $a$ and LHD at $b$ both finite.
Differentiability implies continuity: if $f^{'} (a)$ exists, then $f$ is continuous at $a$ . The converse is false.
Geometrical meaning: $f^{'} (a)$ is the slope of the tangent to the graph of $f$ at $(a, f (a))$ .

1. Definition of the Derivative

The derivative of

f

x

f^{'} (x) = Δ x \to 0 lim \frac{Δ y}{Δ x} = Δ x \to 0 lim \frac{f ( x + Δ x ) - f ( x )}{Δ x},

provided the limit exists finitely. Equivalent notations:

\frac{d y}{d x}

y^{'}

f^{'} (x)

For a two-sided derivative to exist at $x = a$ , the limit must give the same finite value from both sides of $a$ .

1.1 Left-hand derivative (LHD)

f_{-}^{'} (a) = Δ x \to 0^{-} lim \frac{f ( a + Δ x ) - f ( a )}{Δ x} = h \to 0^{+} lim \frac{f ( a - h ) - f ( a )}{- h} .

1.2 Right-hand derivative (RHD)

f_{+}^{'} (a) = Δ x \to 0^{+} lim \frac{f ( a + Δ x ) - f ( a )}{Δ x} = h \to 0^{+} lim \frac{f ( a + h ) - f ( a )}{h} .

Differentiability criterion:

f^{'} (a)

exists iff both LHD and RHD exist, are finite, and are equal. If LHD

\neq =

RHD, or either is infinite, or either fails to exist,

f

is not differentiable at

a

2. Geometrical Meaning

The difference quotient $\frac{f ( x + Δ x ) - f ( x )}{Δ x}$ is the slope of the secant joining the points $(x, f (x))$ and $(x + Δ x, f (x + Δ x))$ on the graph. As $Δ x \to 0$ , the second point slides toward the first, and the secant rotates into the tangent at $(x, f (x))$ . So $f^{'} (x)$ is the slope of the tangent.

Figure 1: As

Δ x \to 0

, the secant through

(x, f (x))

and

(x + Δ x, f (x + Δ x))

rotates into the tangent at

(x, f (x))

. Its slope is

f^{'} (x)

3. Differentiability from First Principle

Solved Example 1

Check whether

f (x) = ∣ x ∣

is differentiable at

x = 0

Solution:

LHD at $0$ :

f_{-}^{'} (0) = h \to 0^{+} lim \frac{f ( - h ) - f ( 0 )}{- h} = h \to 0^{+} lim \frac{h - 0}{- h} = - 1.

RHD at $0$ :

f_{+}^{'} (0) = h \to 0^{+} lim \frac{f ( h ) - f ( 0 )}{h} = h \to 0^{+} lim \frac{h - 0}{h} = 1.

Since LHD $= - 1 \neq = 1 =$ RHD, $∣ x ∣$ is not differentiable at $x = 0$ . Geometrically this is the corner at the origin.

Solved Example 2

Show that

f (x) = {x sin (1/ x), x \neq = 0 0, x = 0

is continuous but not differentiable at

x = 0

Solution:

Continuity: $∣ x sin (1/ x) ∣ \leq ∣ x ∣ \to 0$ , so $lim_{x \to 0} f (x) = 0 = f (0)$ , hence continuous.

Differentiability:

f^{'} (0) = h \to 0 lim \frac{f ( h ) - f ( 0 )}{h} = h \to 0 lim \frac{h sin ( 1/ h ) - 0}{h} = h \to 0 lim sin (1/ h) .

This limit does not exist (oscillates between $- 1$ and $1$ ). So $f$ is not differentiable at $0$ .

4. Reasons for Non-Differentiability

Failure of differentiability at $x = a$ happens in one of four ways.

Figure 2: Corner (

∣ x ∣

), cusp (

x^{2/3}

), vertical tangent (

3 x

), and jump. All four break differentiability - the first three even while continuity holds.

Four failure modes

Corner: LHD and RHD both exist and are finite, but LHD $\neq =$ RHD. Example: $∣ x ∣$ at $x = 0$ .
Cusp: the slope of secants approaches $+ \infty$ from one side and $- \infty$ from the other. Example: $x^{2/3}$ at $x = 0$ .
Vertical tangent: the slope of secants approaches $\pm \infty$ from both sides. Example: $3 x$ at $x = 0$ .
Discontinuity: the function itself is discontinuous at $a$ , which automatically forces non-differentiability.

5. Relation Between Differentiability and Continuity

Key theorem. If

f^{'} (a)

exists (finitely), then

f

is continuous at

a

. The converse is false: continuity does not imply differentiability.

Sketch of proof. If $f^{'} (a)$ exists, then $lim_{h \to 0} (f (a + h) - f (a)) = lim_{h \to 0} h \cdot \frac{f ( a + h ) - f ( a )}{h} = 0 \cdot f^{'} (a) = 0$ , so $lim_{h \to 0} f (a + h) = f (a)$ , meaning $f$ is continuous at $a$ .

Contrapositive form: if $f$ is discontinuous at $a$ , then $f$ cannot be differentiable at $a$ . This is the fastest way to rule out differentiability.

Situation	Continuous?	Differentiable?
Polynomial at any point	Yes	Yes
$∣ x ∣$ at $x = 0$	Yes	No (corner)
$3 x$ at $x = 0$	Yes	No (vertical tangent)
$x sin (1/ x)$ (with $f (0) = 0$ ) at $0$	Yes	No (oscillation of secants)
$[x]$ at $x = 2$	No	No
$x^{2} sin (1/ x)$ (with $f (0) = 0$ ) at $0$	Yes	Yes, $f^{'} (0) = 0$ (but $f^{'}$ not continuous at $0$ )

Fun fact. Differentiability does not imply that the derivative is continuous. The function

x^{2} sin (1/ x)

(with

f (0) = 0

) is differentiable everywhere including at

0

, but

f^{'}

is discontinuous at

0

. There also exist functions like the Weierstrass function that are continuous everywhere but differentiable nowhere.

6. Algebra of Differentiable Functions

If $f$ and $g$ are both differentiable at $x = a$ , then

$f \pm g$ , $f \cdot g$ are differentiable at $a$ .
$f / g$ is differentiable at $a$ , provided $g (a) \neq = 0$ .

Mixed cases - watch out:

Diff + non-diff: if $f$ is differentiable at $a$ and $g$ is not, then $f + g$ is not differentiable at $a$ . (Sum of a smooth function and a corner keeps the corner.)
Non-diff + non-diff: can be differentiable. Example: $f (x) = ∣ x ∣$ and $g (x) = - ∣ x ∣$ ; both non-differentiable at $0$ , but $f + g \equiv 0$ is smooth.
Product diff × non-diff: can be differentiable. Example: $f (x) = x$ and $g (x) = ∣ x ∣$ ; the product $x ∣ x ∣$ is differentiable at $0$ (both LHD and RHD equal $0$ ).

7. Differentiability Involving Functional Equations

Some problems specify $f$ via an equation like $f (x + y) = f (x) + f (y)$ and give one value or a derivative condition. Standard technique: use the functional equation to compute $f^{'} (x)$ from first principle, then integrate.

Solved Example 3

Let

f

satisfy

f (x + f (y)) = f (x) + y

for all

x, y \in R

. Find

f (0)

. Given further that

f (h) = h

for

0 < h < δ

(some positive

δ

), find

f^{'} (x)

and

f (x)

Solution:

Find $f (0)$ : put $x = 0$ , $y = 0$ : $f (0 + f (0)) = f (0) + 0$ , so $f (f (0)) = f (0)$ . Now put $y = 0$ : $f (x + f (0)) = f (x)$ , meaning $f (0)$ is a period of $f$ . Combined with the other relation, taking $x = - f (0)$ gives $f (0) = f (- f (0)) + 0$ , and iterating leads to $2 f (0) = f (0)$ , so $f (0) = 0$ .

Find $f^{'} (x)$ : for $0 < h < δ$ , we have $f (h) = h$ . Then

f^{'} (x) = h \to 0^{+} lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0^{+} lim \frac{f ( x + f ( h )) - f ( x )}{h} = h \to 0^{+} lim \frac{f ( x ) + h - f ( x )}{h} = 1.

So $f^{'} (x) = 1$ everywhere, giving $f (x) = x + C$ . Since $f (0) = 0$ , we get $C = 0$ and $f (x) = x$ .

Solved Example 4

Let

f : R \to R

satisfy

f (x + y^{3}) = f (x) + [f (y)]^{3}

for all

x, y \in R

, and

f^{'} (0) \geq 0

. Find

f (10)

Solution:

Setting $x = y = 0$ : $f (0) = f (0) + [f (0)]^{3}$ , so $[f (0)]^{3} = 0$ , hence $f (0) = 0$ .

By definition, $f^{'} (0) = lim_{h \to 0} \frac{f ( h )}{h}$ . Replace $h$ with $h^{1/3}$ and cube:

[f^{'} (0)]^{3} = h \to 0 lim \frac{[ f ( h ^{1/3} ) ] ^{3}}{h} = h \to 0 lim \frac{f ( h ) - f ( 0 )}{h} = f^{'} (0) .

Let $I = f^{'} (0)$ . Then $I^{3} = I$ , so $I \in {- 1, 0, 1}$ . Given $I \geq 0$ : $I = 0$ or $I = 1$ .

Now compute $f^{'} (x)$ using the functional equation with $y$ replaced by $h^{1/3}$ :

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{f ( x + ( h ^{1/3} ) ^{3} ) - f ( x )}{h} = h \to 0 lim \frac{[ f ( h ^{1/3} ) ] ^{3}}{h} = [f^{'} (0)]^{3} = I .

So $f^{'} (x)$ is constant, $f (x) = I x + C$ ; with $f (0) = 0$ : $f (x) = I x$ . Since $I = 0$ or $1$ : $f (x) = 0$ or $f (x) = x$ . Therefore $f (10) \in {0, 10}$ .

8. Max, Min, and Mid Functions

Functions defined as pointwise maxima or minima of several expressions typically have corners where the "active" branch switches. Differentiability at these switch-points must be checked directly.

Solved Example 5

Let

f (x) = min {tan x, cot x}

for

x \in R

. Find the points of non-differentiability of

f

(0, π)

Solution:

On $(0, π /4)$ : $tan x < 1 < cot x$ (for the middle portion), so $min = tan x$ ; on $(π /4, π /2)$ : $cot x < tan x$ , so $min = cot x$ ; at $x = π /4$ they cross. On $(π /2, 3 π /4)$ : $tan x < 0 < cot x$ ? Recheck: $tan x$ is negative and $cot x$ is negative on $(π /2, π)$ ; comparing signs and magnitudes, they cross at $x = 3 π /4$ .

The switch points where the branch changes are $x = π /4$ and $x = 3 π /4$ . At each, the left- and right-side derivatives are different (one branch is $tan$ , the other is $cot$ ), so $f$ is not differentiable at $x = π /4$ and $x = 3 π /4$ in $(0, π)$ .

Additionally, $f$ is discontinuous at $x = π /2$ (where $tan x$ blows up), so it is trivially not differentiable there either.

Solved Example 6

A function is given by

f (x) = mid {1 - x, ∣ x ∣, 2 x + 1}

, where

mid

selects the median of the three expressions. Find

f (x)

and identify its points of non-differentiability.

Solution:

The three lines have pairwise intersections at $x = - \frac{1}{3}$ (where $2 x + 1 = ∣ x ∣$ for $x < 0$ ), $x = 0$ , and $x = \frac{1}{2}$ (where $1 - x = ∣ x ∣$ for $x > 0$ ).

Comparing values in each interval, the median is:

f (x) = ⎩ ⎨ ⎧ ∣ x ∣, x < - \frac{1}{3} 2 x + 1, - \frac{1}{3} < x < 0 1 - x, 0 < x < \frac{1}{2} ∣ x ∣, x > \frac{1}{2} .

Non-differentiability occurs at each transition point $- \frac{1}{3}, 0, \frac{1}{2}$ (where the active branch switches and the derivatives from the two sides disagree). $f$ is continuous everywhere in $R$ - all three lines agree at each transition point.

Common Mistakes to Avoid

Watch out

Assuming continuity implies differentiability. The direction only goes one way: differentiable $\Rightarrow$ continuous. $∣ x ∣$ is continuous but not differentiable at $0$ .
Skipping the LHD/RHD check at boundary points of piecewise functions. Always verify LHD $=$ RHD $=$ finite at every piece-boundary.
Confusing "derivative exists" with "derivative is continuous." Differentiability does not imply the derivative is continuous. $x^{2} sin (1/ x)$ has a derivative everywhere but $f^{'}$ is discontinuous at $0$ .
Treating vertical tangents as differentiable. If the limit is $+ \infty$ (or $- \infty$ ), the derivative does not exist finitely; the function is not differentiable there.
Assuming a sum or product of non-differentiable functions is non-differentiable. Counter-examples exist ( $∣ x ∣ + (- ∣ x ∣) = 0$ ).
Applying LHD/RHD formulas with the wrong sign. Recall: LHD uses $lim_{h \to 0^{+}} \frac{f ( a - h ) - f ( a )}{- h}$ (note the minus in the denominator). A sign slip flips your answer.
Using $f^{'} (a) = f (a)$ -style shortcuts. The derivative measures the rate of change; it has nothing to do with the value $f (a)$ except through the difference $f (a + h) - f (a)$ .

Frequently Asked Questions

Q1. What is the difference between LHD, RHD, and $f^{'} (a)$ ?

LHD and RHD are one-sided derivatives - the limit of the difference quotient as $x$ approaches $a$ from the left or right only. $f^{'} (a)$ is the two-sided derivative and exists precisely when LHD $=$ RHD $=$ finite value.

Q2. Does differentiability imply continuity?

Yes. If $f^{'} (a)$ exists finitely, then $f$ is continuous at $a$ . The proof uses $f (a + h) - f (a) = h \cdot \frac{f ( a + h ) - f ( a )}{h}$ ; as $h \to 0$ , the right side tends to $0 \cdot f^{'} (a) = 0$ .

Q3. Does continuity imply differentiability?

No. $∣ x ∣$ is continuous at $0$ but has LHD $= - 1$ , RHD $= 1$ , so no derivative. $3 x$ is continuous at $0$ but the tangent is vertical. Some functions are continuous everywhere yet differentiable nowhere (Weierstrass function).

Q4. How do I check differentiability at a boundary point of a piecewise function?

Compute LHD from the left-piece formula and RHD from the right-piece formula, using the definition. First confirm continuity at the point (LHL $=$ RHL $= f (a)$ ). Then confirm LHD $=$ RHD, both finite.

Q5. What is a corner and what is a cusp?

A corner is a point where LHD and RHD both exist and are finite but unequal (like $∣ x ∣$ at $0$ ). A cusp is a sharper failure where the slopes approach $+ \infty$ and $- \infty$ from the two sides (like $x^{2/3}$ at $0$ ). Both make the function non-differentiable.

Q6. Can a function have a derivative at exactly one point?

Yes. Construct $f (x) = x^{2}$ if $x \in Q$ and $f (x) = - x^{2}$ if $x \in / Q$ . Both formulas give $0$ at $x = 0$ and derivative-quotient tending to $0$ from either sequence type, so $f^{'} (0) = 0$ . Elsewhere, $f$ is not even continuous, so cannot be differentiable.

Q7. How is differentiability tested in JEE problems?

Typical setups: piecewise functions with an unknown parameter (find the value making $f$ both continuous and differentiable), functions defined by functional equations (solve for $f$ ), max/min/mid combinations (find where branches switch), and functions involving $∣ x - a ∣$ or $[x]$ or $sin (1/ x)$ (check LHD/RHD carefully at critical points).

Q8. Is every polynomial differentiable everywhere?

Yes. Every polynomial has a derivative at every real number, and the derivative is again a polynomial. This is why polynomials are the "friendliest" functions in calculus.

Previous year questions on Differentiability Of A Function

11 questions from past papers, each with a step-by-step solution.

Show all 11 questions

JEE Advanced 2024 Paper 1, Mathematics Section 4 Q4

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1. Definition of the Derivative

1.1 Left-hand derivative (LHD)

1.2 Right-hand derivative (RHD)

2. Geometrical Meaning

3. Differentiability from First Principle

4. Reasons for Non-Differentiability

5. Relation Between Differentiability and Continuity

6. Algebra of Differentiable Functions

7. Differentiability Involving Functional Equations

8. Max, Min, and Mid Functions

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between LHD, RHD, and f′(a)?

Q2. Does differentiability imply continuity?

Q3. Does continuity imply differentiability?

Q4. How do I check differentiability at a boundary point of a piecewise function?

Q5. What is a corner and what is a cusp?

Q6. Can a function have a derivative at exactly one point?

Q7. How is differentiability tested in JEE problems?

Q8. Is every polynomial differentiable everywhere?

Previous year questions on Differentiability Of A Function

Ready to master Limits, Continuity And Differentiability?

Q1. What is the difference between LHD, RHD, and $f^{'} (a)$ ?