Every rule in differential calculus is a shortcut derived from one definition: the difference quotient. Master the move between definition and shortcut, and differentiation becomes a single idea you can aim at any problem — even ones you have never seen.
1 · Start from the definition
The derivative at a point is the limit of secant slopes as they shrink to zero: f′(x) = lim h→0 (f(x+h) − f(x))/h. Every formula ahead is this limit, pre-evaluated. Asked to use the definition? Expand, factor, cancel h, then let h → 0.
Step 1. Write the definition: f′(x) = lim h→0 ((x+h)² − x²)/h.
Step 2. Expand the top: (x² + 2xh + h²) − x² = 2xh + h².
Step 3. Factor h out: h(2x + h)/h = 2x + h.
Step 4. Let h → 0: f′(x) = 2x. Every power rule below was proved exactly this way.
2 · Constant multiples, sums and differences
Differentiate part by part and constants ride along for the ride: (f + g)′ = f′ + g′, (f − g)′ = f′ − g′, and (k·f)′ = k·f′ for any constant k. So d/dx(3x⁴ − 5x² + 7x − 9) = 12x³ − 10x + 7 — sweep left to right, term by term; constant factors stay put.
3 · Product and quotient rules
Products do not differentiate factor-by-factor. Instead: (uv)′ = u′v + uv′ — the derivative of the first times the untouched second, plus the untouched first times the derivative of the second. Subtraction appears nowhere in a product rule.
For a quotient, the pattern is low d-high minus high d-low, over the square of what is below: (u/v)′ = (u′v − uv′)/v². The order u′v − uv′ is not symmetric — swap the terms and you flip the sign. A reliable habit is to always start with u′v.
4 · Chain rule — functions inside functions
Differentiate the outside while keeping the inside frozen, then multiply by the derivative of the inside: (f(g(x)))′ = f′(g(x))·g′(x). For y = (3x² + 1)⁵ the outer power contributes 5(3x² + 1)⁴ and the inner contributes 6x, giving y′ = 30x(3x² + 1)⁴.
Differentiate y = (x² + 1)³√(5x − 2).
Step 1. Name the two chunks: u = (x² + 1)³ and v = √(5x − 2). This is a product whose factors are each chain-rule composites.
Step 2. u′ = 3(x² + 1)²·2x = 6x(x² + 1)². And v′ = 5/(2√(5x − 2)), because d/dx √(5x − 2) = ½(5x − 2)⁻¹ᐟ² × 5.
Step 3. Apply the product rule: y′ = 6x(x² + 1)²√(5x − 2) + (x² + 1)³·5/(2√(5x − 2)).
Step 4. Clean up by factoring (x² + 1)²/(2√(5x − 2)): y′ = (x² + 1)²(65x² − 24x + 5)/(2√(5x − 2)).
5 · Tangent lines from the derivative
The derivative at a is the slope of the tangent line there, so the tangent line through (a, f(a)) is y − f(a) = f′(a)(x − a). For y = x² − 3x at a = 3: the point is (3, 0), slope f′(3) = 3, line y = 3x − 9. One derivative buys the whole line.
6 · Velocity and acceleration
If s(t) is position, then velocity is v = s′ and acceleration is a = v′ = s″. Every kinematics question is just two derivatives apart. Near Earth's surface gravity gives a constant acceleration a = −9.8 m/s² (sign depends on your up/down convention), and s(t) = s₀ + v₀t + ½at² is the familiar free-fall formula in disguise.
Example: s(t) = t³ − 6t² + 9t metres. Then v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3), a(t) = 6t − 12. At t = 2 the body moves at −3 m/s with zero acceleration; at t = 3, a = +6 m/s².
7 · From the graph of f to the graph of f′
Sketching f′ without a formula: where f rises, f′ is positive; where f falls, f′ is negative; at peaks and valleys f′ = 0; steep stretches of f mean large |f′|. At a corner or cusp, f′ fails to exist — mark it. For piecewise functions, differentiate each piece, then examine the join separately by hand: if both one-sided derivatives agree, f is differentiable there.
Two classic disguises. Piecewise: f(x) = x² for x ≤ 1 and 2x − 1 for x > 1 — both pieces give slope 2 at x = 1, so f′(1) = 2. Limits: lim x→3 (x² − 9)/(x − 3) is exactly the difference-quotient definition of (x²)′ at x = 3, so the answer is 6 — no factoring required once you recognise the pattern (f(x) − f(a))/(x − a).
The chain rule fails silently. Differentiate the outside, keep the inside frozen, then multiply by the inside's derivative. Half of all lost marks on differentiation questions come from that missing final factor — and in quotients, u′v − uv′ is not symmetric: swap the order and the sign flips.
f′(x) = lim h→0 (f(x+h) − f(x))/h
(k f)′ = k f′, (f ± g)′ = f′ ± g′
(uv)′ = u′v + uv′
(u/v)′ = (u′v − uv′)/v²
(f(g(x)))′ = f′(g(x))·g′(x)
(xⁿ)′ = n xⁿ⁻¹, tangent line: y − f(a) = f′(a)(x − a)
v = s′, a = v′ = s″, gravity: a = −9.8 m/s²
Common mistakes
- Differentiating (3x + 1)⁵ as 5(3x + 1)⁴ and stopping — the inner factor 3 is missing.
- Swapping the quotient-rule order to uv′ − u′v, which flips the sign of the answer.
- Substituting the x-value into f before differentiating; differentiate first, evaluate second.
- Skipping the join point of a piecewise function — the outer pieces can agree, or spectacularly disagree, only at the seam.
- Copying the shape of f instead of its slopes when sketching f′ — f′ tracks steepness, not height.
- Writing the tangent line with the wrong anchor point: it must pass through (a, f(a)), not (0, 0).