Continuity and differentiability are fundamental concepts in calculus. A function that can be drawn without lifting your pen is continuous; a function that has a well-defined slope at every point is differentiable. Differentiability is a stronger condition — every differentiable function is continuous, but not every continuous function is differentiable.
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Key Concepts
- 1.Continuity at a Point: A function f is continuous at x = a if:
- 2.f(a) is defined.
- 3.lim_{x → a} f(x) exists.
- 4.lim_{x → a} f(x) = f(a).
Left-hand limit (LHL): lim_{x → a^-} f(x); Right-hand limit (RHL): lim_{x → a^+} f(x).
A function is continuous at a if LHL = RHL = f(a).
Continuity on an interval: f is continuous on [a, b] if it is continuous at every point in (a, b), right-continuous at a, and left-continuous at b.
Differentiability at a Point: f is differentiable at x = a if:
f'(a) = lim_{h → 0} [f(a+h) - f(a)] / h exists (and LHD = RHD).
- Standard Derivatives:
- d/dx(xn) = n · xn-1
- d/dx(ex) = ex
- d/dx(ln x) = 1/x
- d/dx(sin x) = cos x; d/dx(cos x) = -sin x
- d/dx(tan x) = sec2 x; d/dx(cot x) = -cosec2 x
- d/dx(sec x) = sec x tan x; d/dx(cosec x) = -cosec x cot x
- d/dx(sin-1 x) = 1/√(1-x2); d/dx(cos-1 x) = -1/√(1-x2)
- d/dx(tan-1 x) = 1/(1+x2)
Chain Rule: If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x).
Product Rule: d/dx[u · v] = u'v + uv'.
Quotient Rule: d/dx[u/v] = (u'v - uv') / v2.
Implicit Differentiation: Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
Logarithmic Differentiation: Useful when functions are of the form y = [f(x)]g(x). Take ln on both sides, differentiate implicitly.
Parametric Differentiation: If x = f(t) and y = g(t), then dy/dx = (dy/dt)/(dx/dt).
Second Derivative: d2y/dx2 = d/dx(dy/dx). Measures rate of change of slope.
Rolle's Theorem: If f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then there exists c in (a,b) with f'(c) = 0.
Mean Value Theorem (MVT): If f is continuous on [a,b] and differentiable on (a,b), then there exists c in (a,b) with f'(c) = [f(b) - f(a)] / (b - a).
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Worked Examples
Discuss continuity of f(x) = |x| at x = 0.
LHL = lim_{x → 0^-} (-x) = 0. RHL = lim_{x → 0^+} (x) = 0. f(0) = 0.
LHL = RHL = f(0), so f is continuous at x = 0.
Check differentiability of f(x) = |x| at x = 0.
LHD = lim_{h → 0^-} (|0+h| - |0|)/h = lim_{h → 0^-} |h|/h = -1.
RHD = lim_{h → 0^+} |h|/h = 1.
LHD is not equal to RHD, so f(x) = |x| is NOT differentiable at x = 0.
Differentiate y = sin(x2).
Using chain rule: dy/dx = cos(x2) · 2x = 2x cos(x2).
Find dy/dx if x2 + y2 = 25 (implicit differentiation).
Differentiating both sides: 2x + 2y(dy/dx) = 0. So dy/dx = -x/y.
Differentiate y = xx.
Take ln: ln y = x ln x. Differentiate: (1/y)(dy/dx) = ln x + 1. So dy/dx = xx(ln x + 1).
If x = a cos t and y = a sin t, find dy/dx.
dx/dt = -a sin t. dy/dt = a cos t. dy/dx = (a cos t)/(-a sin t) = -cot t.
Verify Rolle's Theorem for f(x) = x2 - 4x + 3 on [1, 3].
f(1) = 1 - 4 + 3 = 0. f(3) = 9 - 12 + 3 = 0. f(1) = f(3). f is continuous and differentiable.
f'(x) = 2x - 4 = 0 => x = 2 in (1, 3). Rolle's theorem is verified.
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Common mistakes
- Assuming every continuous function is differentiable — f(x) = |x| is the classic counterexample.
- Forgetting to apply the chain rule when differentiating composite functions.
- Not checking all three conditions for continuity (existence of f(a), limit, and equality).
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Summary
Continuity requires no breaks; differentiability requires a unique, finite tangent slope. The chain rule, product rule, and implicit differentiation are core techniques. Rolle's theorem and MVT connect function values to derivatives.