JEE Mains Maths · Differentiation

Differentiability of Piecewise Functions

Testing whether a function built from pieces, a composite or a running maximum is differentiable, and finding a derivative at one point from the limit that defines it.

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Why this matters

Eighteen PYQs, sixteen of them multiple choice, and one from 2026. Nine make two pieces meet with matching slopes, or test whether they do; five handle a composite g(f(x)) or a running maximum or minimum; four study functions like x² sin(1/x) near 0, or find f′(0) from the limit that defines it. Three ideas cover the page.

Concept 1 of 3: Where two pieces join

A piecewise function is differentiable at a join exactly when the two pieces meet there and meet with the same slope. That gives two equations, so two unknown constants can be fixed. Continuity alone is not enough, since a corner is continuous. Equal slopes alone are not enough either: two pieces with the same slope at different heights make a jump.

Definition

Continuity at $a$ : left value = right value = $f(a)$ .
Differentiability at $a$ : continuity and left slope = right slope.
For formula pieces, each one-sided slope is that piece's derivative at $a$ .
A piece $\int_0^xg(t)\,dt$ has slope $g(x)$ .

Smooth join at x = a

f_1(a)=f_2(a)\quad\text{and}\quad f_1'(a)=f_2'(a)

Worked example

Find

a

and

b

so that

f(x)=x^2+a

for

x<2

and

f(x)=bx+1

for

x\ge2

is differentiable.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q66Moderate

Example 1 · Differentiation · Differentiability of Piecewise Functions

Let

\alpha,\beta \in R

be such that the function

f\left( x \right)=\left\{ \begin{matrix} 2\alpha\left( x^{2}- 2 \right)+ 2\beta x & ,x < 1 \\ \left( \alpha + 3 \right)x +\left( \alpha - \beta \right) & ,x \geq 1 \end{matrix} \right.

be differentiable at all

x \in R

. Then

34(\alpha + \beta)

is equal to

Continuity comes first

Matching slopes is not enough.

x

for

x<0

and

x+1

for

x\ge0

have slope 1 on both sides but jump at 0, so the function is not differentiable there. Write both equations.

Drill 8 more on where two pieces join

Concept 2 of 3: Composites and running maxima

For

g(f(x))

, trouble can come only from points where

f

is not differentiable, or where

f(x)

lands on a point where

g

is not. Find those

x

, write

g(f(x))

as explicit pieces near each one, and test the join. A function like

\max\{h(t):t\le x\}

is the largest value so far: it follows

h

while

h

rises and stays flat after a peak.

Definition

$g\circ f$ can fail only where $f$ fails or where $f(x)$ hits a bad point of $g$ .
Write $g(f(x))$ in pieces near each such point, then test the join.
$\max\{h(t):t\le x\}$ follows $h$ while $h$ rises and stays flat after a peak.
$\min\{h(t):t\le x\}$ follows $h$ while $h$ falls and stays flat after a trough.

Chain rule at a point

(g\circ f)'(a)=g'\big(f(a)\big)\,f'(a)\quad\text{when both derivatives exist}

Worked example

f(x)=\max\{t^2:-1\le t\le x\}

for

x\ge-1

. Where in

(-1,\infty)

f

not differentiable?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 11 April 2023 · Q170Moderate

Example 2 · Differentiation · Differentiability of Piecewise Functions

Let

f

and

g

be two functions defined by

f(x) =\left\{ \begin{matrix} x + 1, & x < 0 \\ |x - 1|, & x \geq 0 \end{matrix} \right.

and

g(x) =\left\{ \begin{matrix} x + 1, & x < 0 \\ 1, & x \geq 0 \end{matrix} \right.

. Then (gof) (x) is

A bad inner function can be smoothed out

|x|

has a corner at 0, but

|x|^2=x^2

is smooth. Test

g\circ f

itself at each suspect point; do not assume it fails because

f

does.

Drill 4 more on composites and running maxima

Concept 3 of 3: The derivative from its definition

When a function has one formula away from a point and a separate value at it, the derivative there must come from the limit. For

x^2\sin\frac1x

with value 0 at 0, the difference quotient is

h\sin\frac1h\to0

, so

f'(0)=0

; yet

f'(x)=2x\sin\frac1x-\cos\frac1x

has no limit at 0. The derivative exists everywhere but is not continuous. The limit also saves work: if

f(0)=0

, then

f'(0)=\lim\frac{f(x)}{x}

Definition

$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ .
$x^n\sin\frac1x$ with value 0 at 0: continuous for $n\ge1$ , differentiable at 0 for $n\ge2$ , with $f'$ continuous at 0 for $n\ge3$ .
If $f(0)=0$ , then $f'(0)=\lim_{x\to0}\frac{f(x)}{x}$ .

Derivative at a point

f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}

Worked example

f(x)=x^2\sin\frac1x

for

x\ne0

f(0)=0

. Find

f'(0)

, and decide whether

f'

is continuous at 0.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q79Moderate

Example 3 · Differentiation · Differentiability of Piecewise Functions

Suppose

f(x) =\frac{\left( 2^{x}+2^{- x} \right)\tan x\sqrt{\tan^{- 1}\left( x^{2}- x + 1 \right)}}{\left( 7x^{2}+ 3x + 1 \right)^{3}}

Then the value of

f^{'}(0)

is equal to

f′ can exist without being continuous

For

x^2\sin\frac1x

, the formula for

f'(x)

has no limit at 0, but

f'(0)=0

from the definition. Never decide differentiability at a point by taking the limit of the derivative formula.

Drill 3 more on the derivative from its definition

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Where two pieces join
Smooth join at x = a
$f_1(a)=f_2(a)\quad\text{and}\quad f_1'(a)=f_2'(a)$
Composites and running maxima
Chain rule at a point
$(g\circ f)'(a)=g'\big(f(a)\big)\,f'(a)\quad\text{when both derivatives exist}$
The derivative from its definition
Derivative at a point
$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$

Watch out for (3)

Continuity comes first
→ Where two pieces join
A bad inner function can be smoothed out
→ Composites and running maxima
f′ can exist without being continuous
→ The derivative from its definition

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