What is continuity?
A function f is continuous at x = a when three things match:
- the left-hand limit (coming from values just smaller than a),
- the right-hand limit (coming from values just bigger than a),
- the actual value f(a).
In short: lim(x→a) f(x) = f(a). If any one is missing or different, the graph has a break (a hole or a jump) at a.
Polynomials, sin x, cos x, eˣ, and ln x (for x > 0) are continuous wherever they are defined. Sums, differences, products and quotients (denominator not zero) of continuous functions are continuous too.
Differentiable means continuous, not the other way
If a function has a derivative at a, it must be continuous at a. But |x| is continuous at 0 and still has no derivative there: the graph has a sharp corner, so the left slope (−1) and right slope (+1) do not match.
Derivative and the chain rule
The derivative f′(x) is the slope of the tangent at x: f′(x) = lim(h→0) [f(x + h) − f(x)] / h.
Chain rule (a function inside a function): if y = f(u) and u = g(x), then dy/dx = (dy/du) × (du/dx).
Example: y = (3x + 1)⁵. Outer: u⁵ → 5u⁴. Inner: 3x + 1 → 3. So dy/dx = 5(3x + 1)⁴ × 3 = 15(3x + 1)⁴.
Derivatives of inverse trigonometric functions
- d/dx (sin⁻¹x) = 1/√(1 − x²), for −1 < x < 1
- d/dx (cos⁻¹x) = −1/√(1 − x²)
- d/dx (tan⁻¹x) = 1/(1 + x²)
- d/dx (cot⁻¹x) = −1/(1 + x²)
- d/dx (sec⁻¹x) = 1/(|x|√(x² − 1)), d/dx (cosec⁻¹x) = −1/(|x|√(x² − 1))
Why tan⁻¹x: put y = tan⁻¹x, so tan y = x. Differentiate: sec²y · dy/dx = 1, so dy/dx = 1/sec²y = 1/(1 + tan²y) = 1/(1 + x²).
Tip: a smart substitution like x = tan θ often turns a messy expression into a simple one before differentiating.
Implicit functions
Sometimes x and y are mixed together, like x² + y² = 9 or x³ + y³ = 3xy. We do not need to solve for y. Differentiate every term with respect to x, and whenever you differentiate something with y, multiply by dy/dx (chain rule). Then collect the dy/dx terms.
x² + y² = 9 → 2x + 2y·dy/dx = 0 → dy/dx = −x/y.
Exponential and logarithmic functions
- d/dx (eˣ) = eˣ — its slope equals its height.
- d/dx (ln x) = 1/x, for x > 0.
- d/dx (aˣ) = aˣ ln a.
- d/dx (logₐx) = 1/(x ln a).
With the chain rule: d/dx (e^(3x)) = 3e^(3x), d/dx ln(sin x) = cos x / sin x = cot x.
Logarithmic differentiation
Use it when the power itself has x (like xˣ or (sin x)^x), or for a long product/quotient.
- Take ln of both sides: ln y = x ln x.
- Differentiate: (1/y) dy/dx = ln x + 1.
- Multiply by y: dy/dx = xˣ(1 + ln x).
Functions in parametric form
Sometimes x and y both depend on a third variable t (a parameter), like x = t², y = 2t. Then
dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0.
For a circle x = r cos t, y = r sin t: dy/dx = (r cos t)/(−r sin t) = −cot t.
Second order derivative
Differentiate once more: d²y/dx² = d/dx (dy/dx), also written f″(x) or y₂.
It tells how the slope changes. f″ > 0: the slope is growing, the curve bends up like a bowl. f″ < 0: bends down like a cap.
Example: y = x³ → y′ = 3x² → y″ = 6x.
For parametric forms: d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt). Do not simply divide d²y/dt² by d²x/dt².
Key formulas and definitions
- Continuity at a: lim(x→a⁻) f = lim(x→a⁺) f = f(a)
- Chain rule: dy/dx = (dy/du)(du/dx)
- d/dx sin⁻¹x = 1/√(1 − x²), d/dx tan⁻¹x = 1/(1 + x²)
- d/dx eˣ = eˣ, d/dx ln x = 1/x, d/dx aˣ = aˣ ln a
- Parametric: dy/dx = (dy/dt)/(dx/dt)
- Second derivative: d²y/dx² = d/dx (dy/dx)
Worked examples
1. Is f(x) = 2x + 3 for x ≤ 1 and f(x) = 6 − x for x > 1 continuous at x = 1?
Left limit: 2(1) + 3 = 5. Right limit: 6 − 1 = 5. Value f(1) = 5. All three equal, so f is continuous at x = 1.
2. Find k so that f(x) = kx² for x ≤ 2 and f(x) = 3 for x > 2 is continuous at x = 2.
Left limit = value = 4k. Right limit = 3. For continuity 4k = 3, so k = 3/4.
3. Differentiate y = sin(x² + 5).
Outer sin u → cos u; inner x² + 5 → 2x. dy/dx = cos(x² + 5) × 2x = 2x cos(x² + 5).
4. Differentiate y = tan⁻¹(2x/(1 − x²)) for −1 < x < 1.
Put x = tan θ. Then 2x/(1 − x²) = tan 2θ, so y = 2θ = 2 tan⁻¹x. dy/dx = 2/(1 + x²).
5. Find dy/dx if x² + xy + y² = 7.
2x + (y + x·dy/dx) + 2y·dy/dx = 0. So dy/dx (x + 2y) = −(2x + y), giving dy/dx = −(2x + y)/(x + 2y).
6. Differentiate y = e^(3x) · ln x.
Product rule: dy/dx = 3e^(3x) ln x + e^(3x)(1/x) = e^(3x)(3 ln x + 1/x).
7. Differentiate y = xˣ (x > 0).
ln y = x ln x. (1/y) dy/dx = ln x + 1. dy/dx = xˣ(1 + ln x).
8. If x = a cos t and y = a sin t, find dy/dx and d²y/dx².
dx/dt = −a sin t, dy/dt = a cos t, so dy/dx = −cot t. d/dt(−cot t) = cosec²t. d²y/dx² = cosec²t ÷ (−a sin t) = −1/(a sin³t).
9. If y = 3e^(2x) + 2e^(−x), show that y″ − y′ − 2y = 0.
y′ = 6e^(2x) − 2e^(−x); y″ = 12e^(2x) + 2e^(−x). y″ − y′ − 2y = (12 − 6 − 6)e^(2x) + (2 + 2 − 4)e^(−x) = 0.
Common mistakes
- Forgetting the inner derivative in the chain rule: d/dx sin(3x) is 3cos(3x), not cos(3x).
- In implicit differentiation, writing d/dx(y²) = 2y instead of 2y·dy/dx.
- Thinking every continuous function is differentiable. |x| is continuous at 0 but has a corner, so no derivative there.
- For parametric second derivatives, dividing d²y/dt² by d²x/dt². Instead differentiate dy/dx with respect to t and divide by dx/dt.