Continuity
Informally, a function is continuous if you can draw its graph without lifting your pen.
Formally, f is continuous at x = c if
lim (x → c) f(x) = f(c)
That needs three things: f(c) is defined, the limit exists (left-hand limit = right-hand limit), and the limit equals f(c).
f is continuous on an interval if it is continuous at every point of it.
Example. f(x) = { x + 2 for x ≤ 1; 4 − x for x > 1 }. At x = 1: LHL = 3, RHL = 3, f(1) = 3. So f is continuous at 1.
Example. f(x) = { x for x < 0; x + 1 for x ≥ 0 }. At x = 0, LHL = 0 but RHL = 1, so f is discontinuous there (a jump).
Finding an unknown constant. For f(x) = { kx² for x ≤ 2; 3 for x > 2 } to be continuous at 2, we need 4k = 3, so k = 3/4.
Continuous functions you can rely on: polynomials, sin x, cos x, eˣ (everywhere); rational functions, tan x and log x (everywhere in their domains); |x| (everywhere, including 0). Sums, differences, products, quotients (where the denominator is non-zero) and compositions of continuous functions are continuous.
Differentiability
f is differentiable at c if the derivative
f′(c) = lim (h → 0) [f(c + h) − f(c)]/h
exists, which means the left-hand derivative and right-hand derivative are equal.
Every differentiable function is continuous, but a continuous function need not be differentiable.
Example. f(x) = |x| is continuous at 0, but the left derivative is −1 and the right derivative is +1. The graph has a sharp corner, so it is not differentiable at 0.
The chain rule
For a composite function y = f(g(x)):
dy/dx = f′(g(x)) × g′(x), or dy/dx = (dy/du)(du/dx) with u = g(x).
Differentiate the outer function, keep the inside the same, then multiply by the derivative of the inside.
Example. d/dx sin(x²) = cos(x²) × 2x = 2x cos(x²).
Example. d/dx (3x + 1)⁵ = 5(3x + 1)⁴ × 3 = 15(3x + 1)⁴.
Standard derivatives
f(x) | f′(x) |
|---|---|
eˣ | eˣ |
log x (natural log) | 1/x, x > 0 |
aˣ | aˣ log a |
sin⁻¹ x | 1 ⁄ √(1 − x²) |
cos⁻¹ x | −1 ⁄ √(1 − x²) |
tan⁻¹ x | 1 ⁄ (1 + x²) |
sec x | sec x tan x |
cosec x | −cosec x cot x |
cot x | −cosec² x |
Implicit differentiation
When y is not given explicitly as a function of x, differentiate both sides with respect to x, treating y as a function of x (so d/dx of y² is 2y dy/dx), then solve for dy/dx.
Example. x² + y² = 25. Differentiating: 2x + 2y (dy/dx) = 0, so dy/dx = −x/y.
Example. sin y = xy. Differentiating: cos y (dy/dx) = y + x (dy/dx), so dy/dx = y ⁄ (cos y − x).
Logarithmic differentiation
Useful when the variable is in both the base and the exponent, or for long products and quotients. Take logs first, then differentiate.
Example. y = xˣ. Then log y = x log x. Differentiating: (1/y)(dy/dx) = log x + 1, so dy/dx = xˣ(1 + log x).
Parametric differentiation
If x = f(t) and y = g(t), then
dy/dx = (dy/dt) ⁄ (dx/dt)
Example. x = a cos θ, y = a sin θ. dy/dx = (a cos θ)/(−a sin θ) = −cot θ.
Second order derivatives
The derivative of dy/dx is the second derivative, written d²y/dx² or y″.
Example. If y = A sin x + B cos x, then y″ = −A sin x − B cos x = −y, so y″ + y = 0.
Mean value theorems
Rolle's theorem. If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there is at least one c in (a, b) with f′(c) = 0. Geometrically, somewhere the tangent is horizontal.
Lagrange's mean value theorem. If f is continuous on [a, b] and differentiable on (a, b), there is at least one c in (a, b) with
f′(c) = [f(b) − f(a)] ⁄ (b − a)
So somewhere the tangent is parallel to the chord joining the end points.
Example. For f(x) = x² on [2, 4]: [f(4) − f(2)]/(4 − 2) = (16 − 4)/2 = 6. Setting f′(c) = 2c = 6 gives c = 3, which lies in (2, 4).
Summary
Continuity at c: the limit exists and equals f(c).
Differentiable ⇒ continuous, but not the reverse (|x| at 0).
Chain rule: differentiate outer × derivative of inner.
Use implicit, logarithmic or parametric differentiation as the form demands.
Rolle's theorem: f(a) = f(b) ⇒ f′(c) = 0 somewhere; MVT: f′(c) equals the slope of the chord.