Continuity and Differentiability

2.1 Core concepts

• Continuity at a point: A real function f is continuous at c in its domain if lim_{x→c} f(x) = f(c); equivalently, the left-hand limit, right-hand limit, and f(c) all exist and are equal (NCERT §5.2, p. 105). All three conditions are needed; relaxing any one breaks continuity.
• Continuity on domain: f is continuous if it is continuous at every point of its domain; on [a, b], continuity at endpoints uses only one-sided limits — lim_{x→a+} f(x) = f(a) and lim_{x→b−} f(x) = f(b) (NCERT §5.2, p. 107).
• Standard continuous functions established in worked examples: constant function, identity f(x)=x, polynomial, rational (where denominator ≠ 0), |x|, sin x, cos x, and tan x wherever defined (NCERT §5.2, pp. 106–115). These serve as "building blocks" — combinations of them are continuous wherever the algebra is legal.
• Greatest integer function f(x) = [x] is discontinuous at every integer and continuous everywhere else (NCERT §5.2, Example 15, p. 112). The jumps are of size 1 at each integer.
• Algebra of continuous functions (Theorem 1): if f and g are continuous at c, then f ± g, f·g, and f/g (when g(c) ≠ 0) are continuous at c (NCERT §5.2.1, p. 113).
• Composition (Theorem 2): if g is continuous at c and f is continuous at g(c), then f ∘ g is continuous at c (NCERT §5.2.1, p. 115). The composition theorem powers most "is this function continuous?" arguments.
• Derivative definition: f'(c) = lim_{h→0} [f(c+h) − f(c)]/h when this limit exists; f is differentiable at c iff the left and right derivatives are finite and equal (NCERT §5.3, p. 118).
• Theorem 3 (Differentiability ⇒ Continuity): every differentiable function is continuous; the converse fails — f(x) = |x| is continuous but not differentiable at x = 0 (NCERT §5.3, p. 120). This is a one-way implication of singular importance for CUET.
• Algebra of derivatives: (u ± v)' = u' ± v', product rule (uv)' = u'v + uv', quotient rule (u/v)' = (u'v − uv')/v² for v ≠ 0 (NCERT §5.3, p. 119).
• Chain rule (Theorem 4): for f = v ∘ u with t = u(x), df/dx = (dv/dt)·(dt/dx); extended to triple compositions as (dw/ds)(ds/dt)(dt/dx) (NCERT §5.3.1, p. 121). The chain rule is the workhorse of all composite-function derivatives.
• Implicit differentiation: when y is given implicitly by a relation in x and y, differentiate both sides w.r.t. x treating y as a function of x, then solve for dy/dx (NCERT §5.3.2, pp. 122–123).
• Derivatives of inverse trig functions: d/dx(sin⁻¹ x) = 1/√(1 − x²), d/dx(cos⁻¹ x) = −1/√(1 − x²), d/dx(tan⁻¹ x) = 1/(1 + x²); sin⁻¹ and cos⁻¹ derivatives valid on (−1, 1), tan⁻¹ valid on R (NCERT §5.3.3, p. 124).
• Exponential & log derivatives (Theorem 5): d/dx(eˣ) = eˣ and d/dx(log x) = 1/x for x > 0; in this chapter log x denotes natural log (base e) (NCERT §5.4, p. 129). The exponential is the unique function (up to constant multiple) equal to its own derivative.
• Derivative of aˣ (Example 28): d/dx(aˣ) = aˣ log a for positive constant a; obtained by writing aˣ = e^{x log a} or by logarithmic differentiation (NCERT §5.5, p. 131).
• Change-of-base formula: log_a p = (log_b p)/(log_b a); applied in Example 39 to differentiate log₇(log x) as (log(log x))/log 7 (NCERT §5.4, p. 128; Miscellaneous Example, p. 141).
• Logarithmic differentiation: for y = [u(x)]^{v(x)} (with u(x) > 0), take log y = v(x) log u(x) then differentiate; standard technique for xˣ, xsin x, etc. (NCERT §5.5, p. 130). Without taking logs first, the standard power-rule and exponential-rule cannot be applied directly.
• Parametric differentiation: if x = f(t), y = g(t), then dy/dx = (dy/dt)/(dx/dt) = g'(t)/f'(t) whenever f'(t) ≠ 0 (NCERT §5.6, p. 135). Used widely in curves like the cycloid and ellipse.
• Second-order derivative: d²y/dx² = d/dx(dy/dx), also denoted f''(x), y'', y₂, or D²y; higher orders defined analogously (NCERT §5.7, p. 137). Second derivatives feature in concavity, inflection, and physical interpretations like acceleration.
• These derivatives and rules are foundational for Class XII Applications of Derivatives (lemh106), Integrals (lemh107) and beyond — they recur throughout the calculus syllabus.
• The historical thread: continuity (Cauchy, Weierstrass) and differentiability (Newton, Leibniz) were unified into modern analysis through ε–δ definitions. NCERT uses the intuitive limit-based definitions rather than full ε–δ rigour, sufficient for Class XII problem solving.
• A subtle observation: every polynomial is everywhere differentiable; every rational function is differentiable on its domain; every trigonometric function is differentiable on its domain; every exponential is differentiable everywhere on R. Standard functions are well-behaved; trouble arises mainly at piecewise junctions, isolated points (like |x| at 0), or where denominators vanish.
• Newton's and Leibniz's notations both appear: dy/dx (Leibniz) and f'(x) (Lagrange) are the two main systems. Both denote the same derivative; CUET uses Lagrange notation more often in MCQ stems.
• Second-derivative significance: d²y/dx² > 0 ⇒ concave up; < 0 ⇒ concave down; = 0 with change of sign ⇒ inflection point. These ideas explicitly enter the next chapter on Applications of Derivatives.

2.2 Definitions to memorise

Standard derivatives summarised (Summary box, p. 146 and Table 5.3, p. 119):

2.3 Diagrams / processes to remember

• Fig 5.1 (p. 104) – Step function with jump at x = 0: shows a function that cannot be drawn without lifting the pen, motivating the visual idea of discontinuity.
• Fig 5.3 (p. 109) – Graph of f(x) = 1/x: illustrates that left-hand limit is −∞ and right-hand limit is +∞ at x = 0, neither being a real number.
• Fig 5.5 (p. 110) – Piecewise function with isolated point of discontinuity at x = 1: typifies the "jump + redefined value" trap.
• Fig 5.8 (p. 112) – Graph of the greatest integer function: step-like staircase, discontinuous at every integer.
• Figs 5.9–5.11 (pp. 125–128) – Comparison of polynomial growth xⁿ vs exponential 10ˣ, plot of y = bˣ, and y = eˣ and y = ln x as mirror images across y = x.
• Process — check continuity at point c: (i) compute lim_{x→c⁻} f(x); (ii) compute lim_{x→c⁺} f(x); (iii) check both equal; (iv) compare with f(c). If all three agree, f is continuous at c.
• Process — find k so f is continuous: equate the relevant one-sided limit to f(c) and solve for the unknown constant.
• Process — differentiate composite via chain rule: identify outer and inner functions; differentiate outer (keeping inner intact), multiply by derivative of inner.
• Process — logarithmic differentiation: take ln of both sides; differentiate implicitly; solve for dy/dx; multiply by y to express explicitly.
• Process — parametric dy/dx: find dx/dt and dy/dt; divide; ensure dx/dt ≠ 0 at the point of interest.

2.4 Common confusions / NTA trap points

• "Continuous ⇒ differentiable" is false. The standard counter-example is f(x) = |x| at x = 0: left derivative is −1, right derivative is +1, so the function is continuous but not differentiable there (p. 120).
• For piecewise definitions, you must check both that one-sided limits agree and that the common value equals f(c). Example 4 (f(x) = x³ + 3 if x ≠ 0, f(0) = 1) fails the second condition even though the limit exists (p. 107).
• lim_{x→0⁺} 1/x = +∞ and lim_{x→0⁻} 1/x = −∞ — but ±∞ are not real numbers, so these limits do not exist as real numbers (p. 109).
• In this chapter log x means natural log (base e), not base 10. So d/dx(log x) = 1/x, not 1/(x ln 10) (p. 128).
• For d/dx(aˣ) = aˣ log a, the log a is ln a; students wrongly write aˣ or x aˣ⁻¹. The correct derivation uses aˣ = e^{x log a} (Example 28, p. 131).
• The derivative of sin⁻¹ x is only valid on the open interval (−1, 1); at x = ±1 the derivative is not defined because cos y = 0 there (p. 124).
• For y = xˣ, neither the power rule (x · xˣ⁻¹) nor the exponential rule (xˣ log x) is correct in isolation; use logarithmic differentiation to get xˣ(1 + log x).
• Forgetting to differentiate the inner function in chain rule — d/dx[sin(x²)] = 2x cos(x²), not cos(x²).
• Mis-applying quotient rule with sign error — denominator squared is always v², not v.
• Confusing implicit and parametric: implicit has y appearing in a relation with x; parametric has x and y both as functions of a third variable t.
• Forgetting to keep dy/dx on one side when doing implicit differentiation. After differentiating, collect dy/dx terms on LHS and solve.
• Stopping at first derivative when second is asked. Always differentiate again.
• Confusing log_a x with ln x; the conversion is log_a x = ln x / ln a.
• Treating the derivative of a constant as anything other than 0. Constants vanish under differentiation.
• Forgetting that the chain rule applies even to nested compositions of three or more functions; each level contributes a factor.
• Mis-using the quotient rule by reversing numerator and denominator. The correct numerator is u'v − uv', NOT v'u − vu'.
• Skipping the absolute-value in the derivatives of sec⁻¹ x and cosec⁻¹ x: they have |x| in the denominator because the range of these functions is restricted to [0, π] − {π/2} and [−π/2, π/2] − {0} respectively.
• Computing the derivative of (sin x)^x using the power rule alone — this is a "function to function" power, requiring logarithmic differentiation.
• Misinterpreting "find dy/dx" in parametric form as differentiating y w.r.t. t; the correct interpretation is the ratio (dy/dt)/(dx/dt).
• Forgetting that continuity at a single point does not imply continuity on an interval — verifying continuity at one point is necessary but not sufficient.
• Mishandling derivative of inverse trigonometric functions when the input is not x but a function of x (e.g., sin⁻¹(2x − 1) requires chain rule: derivative is 2/√(1 − (2x − 1)²)).
• Treating |x| as differentiable everywhere; it is differentiable on R \ {0} but not at 0 itself, where the corner has slopes ±1.
• Confusing the chain rule's "outer × inner" with reverse order — the outer function's derivative comes first.
• Forgetting that the product rule symbol u'v + uv' is symmetric; some students drop one of the two terms.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 10, p. 110). Is f continuous at x = 1, where f(x) = x + 2 for x ≤ 1 and f(x) = x − 2 for x > 1?

Step 1 — LHL: lim_{x→1⁻}(x + 2) = 3. Step 2 — RHL: lim_{x→1⁺}(x − 2) = −1. Step 3 — compare: LHL ≠ RHL ⇒ discontinuous at x = 1.

Example B (NCERT Example 21, p. 121). Differentiate sin(x²).

Step 1 — let t = x²: y = sin t. Step 2 — chain rule: dy/dx = (dy/dt)(dt/dx) = cos t · 2x. Step 3 — substitute: 2x cos(x²).

Example C (NCERT Example 28, p. 131). Find d/dx (aˣ).

Step 1 — log both sides of y = aˣ: log y = x log a. Step 2 — differentiate: (1/y)(dy/dx) = log a. Step 3 — solve: dy/dx = y log a = aˣ log a.

Example D (NCERT Example 30, p. 133). Find dy/dx for y = xˣ, x > 0.

Step 1 — log both sides: log y = x log x. Step 2 — differentiate: (1/y)(dy/dx) = log x + x(1/x) = 1 + log x. Step 3 — solve: dy/dx = xˣ(1 + log x).

Example E (NCERT Example 32, p. 135). x = at², y = 2at. Find dy/dx.

Step 1 — dx/dt: 2at. Step 2 — dy/dt: 2a. Step 3 — divide: dy/dx = 2a/(2at) = 1/t.

Example F (NCERT Example 35, p. 138). Find d²y/dx² for y = x³ + tan x.

Step 1 — first derivative: dy/dx = 3x² + sec² x. Step 2 — differentiate term-by-term: d/dx(3x²) = 6x; d/dx(sec² x) = 2 sec x · (sec x tan x) = 2 sec² x tan x. Step 3 — combine: d²y/dx² = 6x + 2 sec² x tan x.

Example G (Continuity at a junction). Find k so that f(x) = kx² for x ≤ 2 and f(x) = 3 for x > 2 is continuous at x = 2.

Step 1 — LHL: lim_{x→2⁻}(kx²) = 4k. Step 2 — RHL: lim_{x→2⁺}(3) = 3. Step 3 — equate: 4k = 3 ⇒ k = 3/4.