Differentiability: Left and Right Hand Derivatives | NIMCET

Quick answer: f is differentiable at a if $f'(a) = \lim_{h \to 0}\frac{f(a + h) - f(a)}{h}$ exists, that is, the left-hand derivative $\lim_{h \to 0^-}\frac{f(a + h) - f(a)}{h}$ equals the right-hand derivative $\lim_{h \to 0^+}\frac{f(a + h) - f(a)}{h}$ . Differentiability implies continuity, but not the other way round: $\lvert x\rvert$ is continuous at 0 but has LHD = −1 and RHD = 1, so it is not differentiable there.

This continues from continuity of functions.

Derivative from first principles

$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$

Geometrically, f′(a) is the slope of the tangent to y = f(x) at x = a.

Left and right hand derivatives

Derivative	Formula
LHD at a	$\lim_{h \to 0^+}\frac{f(a - h) - f(a)}{-h}$
RHD at a	$\lim_{h \to 0^+}\frac{f(a + h) - f(a)}{h}$
Differentiable	LHD = RHD (finite)

Differentiability and continuity

Statement	True?
Differentiable at a ⇒ continuous at a	Yes
Continuous at a ⇒ differentiable at a	No
Not continuous ⇒ not differentiable	Yes

So when testing differentiability, check continuity first; a discontinuous function cannot be differentiable.

Where differentiability usually fails

Situation	Example
Corner (sharp turn)	$\lvert x\rvert$ at 0
Vertical tangent	$x^{\frac{1}{3}}$ at 0
Discontinuity	$[x]$ at integers
Join of a piecewise function with unequal slopes	$\max(x, x^2)$ at 0 and 1

Standard cases

Function	Not differentiable at
$\lvert x - a\rvert$	$x = a$
$\lvert x\rvert + \lvert x - 1\rvert$	$x = 0, 1$
$x\lvert x\rvert$	Nowhere (differentiable everywhere)
$\lvert\sin x\rvert$	$x = n\pi$
$[x]$	Every integer

A quick rule: $\lvert f(x)\rvert$ can fail to be differentiable only where $f(x) = 0$ , and fails there when $f'(x) \ne 0$ at that point.

Worked examples

Example 1: modulus function

Check $f(x) = \lvert x\rvert$ at $x = 0$ .

RHD $= \lim_{h \to 0^+}\frac{h}{h} = 1$ ; LHD $= \lim_{h \to 0^+}\frac{h}{-h} = -1$ . Not differentiable.

Example 2: x|x|

Check $f(x) = x\lvert x\rvert$ at 0.

RHD $= \lim\frac{h \cdot h}{h} = 0$ and LHD $= \lim\frac{(-h)(h)}{-h} = 0$ . Differentiable, with $f'(0) = 0$ .

Example 3: piecewise constants

$f(x) = x^2$ for $x \le 1$ and $f(x) = ax + b$ for $x > 1$ . Find a, b for differentiability at 1.

Step 1: Continuity: $1 = a + b$ .

Step 2: Equal derivatives: $2 = a$ .

Answer: $a = 2$ , $b = -1$ .

Example 4: counting points

Find the points where $f(x) = \lvert x^2 - 3x + 2\rvert$ is not differentiable.

$x^2 - 3x + 2 = 0$ at $x = 1, 2$ , and the derivative $2x - 3$ is non-zero at both. So f fails at $x = 1$ and $x = 2$ .

Example 5: first principles

Find the derivative of $x^2$ from first principles.

$\lim_{h \to 0}\frac{(x + h)^2 - x^2}{h} = \lim_{h \to 0}(2x + h) = 2x$

Summary table

Idea	Key result
Differentiable	LHD = RHD
Differentiable ⇒ continuous	Always
Continuous ⇒ differentiable	Not always
$\lvert x\rvert$ at 0	LHD −1, RHD 1
Piecewise	Match values and slopes at the join

Exam tips

For piecewise functions, write two equations: one for continuity and one for equal slopes.
Sketching the graph shows corners and jumps at a glance.
$x^n\lvert x\rvert$ with $n \ge 1$ is differentiable at 0.

Next, learn the chain rule and implicit and parametric differentiation, or see the NIMCET Calculus chapter.

Common mistakes

Concluding differentiability from continuity.
Checking only the slope condition at a join and ignoring continuity.
Forgetting the minus sign in the LHD formula.
Assuming every modulus function has a corner at its zero.

Frequently Asked Questions

When is a function differentiable at a point?

When its left-hand and right-hand derivatives at that point exist and are equal.

Is every continuous function differentiable?

No. For example, |x| is continuous at 0 but not differentiable there.

Is every differentiable function continuous?

Yes. Differentiability at a point implies continuity at that point.

Why is |x| not differentiable at 0?

Its left-hand derivative is −1 and its right-hand derivative is 1, so they are unequal.

What is the derivative from first principles?

f′(x) = lim (h → 0) [f(x + h) − f(x)]/h.