10 min readConcept sheet1-mark MCQ and assertion-reason; 2 or 3-mark 'check differentiability' questions; also a JEE favourite

Differentiability and its link with continuity

f is differentiable at c when its left and right derivatives there are equal. Differentiable means continuous, but continuous need not mean differentiable.

Derivative at cf′(c) = lim (h → 0) [f(c + h) − f(c)]/h, if this limit exists

LHD and RHDLet h → 0⁻ for the left-hand derivative and h → 0⁺ for the right-hand derivative; f is differentiable at c when they are equal

Key theoremIf f is differentiable at c, then f is continuous at c. The converse is false.

Standard counter-example|x| is continuous at 0 but not differentiable there: LHD = −1, RHD = 1

What differentiable means

f is differentiable at c if its graph has a single, clear tangent there. In algebra, the left-hand derivative (LHD) and the right-hand derivative (RHD) must exist and be equal. A sharp corner (like the V of |x|) or a break in the graph means there is no derivative at that point. f is differentiable on an interval if it is differentiable at every point of it.

Example 1: a sharp corner

Problem

Show that f(x) = |x − 3| is continuous but not differentiable at x = 3.

Working

f(3) = 0. LHL = lim (x → 3⁻) (3 − x) = 0 and RHL = lim (x → 3⁺) (x − 3) = 0. So f is continuous at 3.
LHD = lim (h → 0⁻) [f(3 + h) − f(3)]/h = lim (h → 0⁻) |h|/h
For h < 0, |h| = −h, so LHD = lim (−h)/h = −1.
RHD = lim (h → 0⁺) |h|/h = lim h/h = 1, since |h| = h for h > 0.
LHD ≠ RHD, so f is not differentiable at 3.

Answer

f is continuous at x = 3 but not differentiable there (a sharp corner).

Example 2: a smooth join

Problem

Check whether f(x) = {x², if x ≤ 1; 2x − 1, if x > 1} is differentiable at x = 1.

Working

f(1) = 1, and RHL = 2(1) − 1 = 1, so f is continuous at 1.
LHD = lim (h → 0⁻) [(1 + h)² − 1]/h = lim (2h + h²)/h = lim (2 + h) = 2
RHD = lim (h → 0⁺) [2(1 + h) − 1 − 1]/h = lim 2h/h = 2
LHD = RHD = 2

Answer

f is differentiable at x = 1, and f′(1) = 2.

Algebra of derivatives and standard results

(u ± v)′ = u′ ± v′
Product rule: (uv)′ = u′v + uv′
Quotient rule: (u/v)′ = (u′v − uv′)/v², where v ≠ 0
d/dx xⁿ = nxⁿ⁻¹; d/dx sin x = cos x; d/dx cos x = −sin x; d/dx tan x = sec²x
d/dx cot x = −cosec²x; d/dx sec x = sec x tan x; d/dx cosec x = −cosec x cot x
Every polynomial is differentiable everywhere.

Continuity vs differentiability

Continuous at cNo break: LHL = RHL = f(c). A sharp corner is allowed.

Differentiable at cNo break and no sharp corner: LHD = RHD. This is the stronger condition.

Differentiable ⇒ continuous. Continuous does not imply differentiable: think of |x| at 0.

In real life

A smooth ghat road has a definite slope at every point. A road with a hairpin corner is still joined up (continuous) but has no single slope at the corner, which is why such bends carry warning signs.

Exam angle

1 marks: “Is f(x) = |x − 3| differentiable at x = 3? Give a reason.” No. LHD = −1 and RHD = 1 are not equal: the graph has a sharp corner at x = 3.

3 marks: “Check whether f(x) = {x³, if x ≤ 1; 3x − 2, if x > 1} is differentiable at x = 1.” Continuous: f(1) = 1 = RHL. LHD = lim [(1 + h)³ − 1]/h = lim (3 + 3h + h²) = 3. RHD = lim [3(1 + h) − 2 − 1]/h = 3. Equal, so f is differentiable at 1 and f′(1) = 3.

2 marks: “Differentiate y = (x² + 1)/(x − 1), x ≠ 1, using the quotient rule.” dy/dx = [2x(x − 1) − (x² + 1)(1)]/(x − 1)² = (x² − 2x − 1)/(x − 1)².

Remember it

Every smooth curve is joined, but not every joined curve is smooth. Differentiable ⇒ continuous, never the reverse.