FSc Part 1 Mathematics Chapter 13 Complete Guide Book — Differentiation

How do you find the gradient of a curve at a single point, when gradient seems to need two points? The book's answer is a limit in disguise. Take two points on the curve, P = (x, f(x)) and Q = (x + δx, f(x + δx)). The line PQ is a secant, and its slope is m_sec = [f(x + δx) − f(x)]/δx — just rise over run. Now let Q slide along the curve towards P, so δx → 0. The secant line settles onto the tangent at P, and its slope settles onto the tangent's gradient: m_tan = lim(δx→0) [f(x + δx) − f(x)]/δx. The whole chapter is this one limit, unpacked.

Example 1. For f(x) = x² − 2 at the point (−1, −1): the difference quotient is [(−1 + δx)² − 2 − (−1)]/δx = (−2δx + (δx)²)/δx = −2 + δx, which tends to −2 as δx → 0. So the gradient is −2, and the tangent line is y − (−1) = −2(x − (−1)), i.e. y = −2x − 3. Check it passes through (−1, −1): −2(−1) − 3 = −1. ✓

The same limit measures rates of change. The average rate of f over [x, x + δx] is [f(x + δx) − f(x)]/δx — the same quotient as the secant slope, read as “change in f per unit change in x”. The instantaneous rate is the limit of that quotient as δx → 0, and we call it the derivative: f′(x) = lim(δx→0) [f(x + δx) − f(x)]/δx. A function that has a derivative at x is differentiable at x. So “gradient of the tangent”, “instantaneous rate of change” and “derivative” are three names for one number.

13.2.1 Velocity. For motion s(t) = t² + t, the average velocities over shrinking intervals ending at t = 3 come out as 9, 8, 7.5, 7.1, 7.01, 7.001 — visibly homing in on 7. So the instantaneous velocity at t = 3 is 7 m/s, and in general v_inst = lim(δt→0) [s(t + δt) − s(t)]/δt = s′(t). Example 2: for s = 4t² + 2t + 1, the average velocity over [2, 5] is (s(5) − s(2))/3 = (111 − 21)/3 = 30 miles/h, while the instantaneous velocity at t = 3 is s′(3) = 8(3) + 2 = 26 miles/h. Average and instantaneous answer different questions — do not swap them.

Finding a derivative straight from the limit definition is called differentiation ab-initio or from first principles, and the book's four steps are worth memorising: (1) write f(x + δx); (2) form the difference f(x + δx) − f(x); (3) divide by δx and simplify — this is where a factor δx must appear and cancel; (4) let δx → 0. If step (3) leaves no δx to cancel, you have made an algebra slip; go back, do not force the limit.

Notations (13.3.1 — ⚠️ the book prints this as “12.3.1”, a numbering slip; read it as 13.3.1): Leibniz dy/dx, Newton ḟ or ẏ, Lagrange f′(x), Euler Df(x). 📌 Book note: dy/dx is a single symbol for the derivative, not a quotient of dy by dx — even though the definition looks like one. The derivative at a point a can also be written f′(a) = lim(x→a) [f(x) − f(a)]/(x − a), which is the same limit with the roles of the two points swapped.

Example 3: (a) a constant c gives [c − c]/δx = 0, so d(c)/dx = 0 — flat lines have zero slope. (b) f(x) = x² gives [(x + δx)² − x²]/δx = 2x + δx → 2x. Example 4: f(x) = √x gives, after rationalising, 1/(√(x + δx) + √x) → 1/(2√x); and f′(a) = 1/(2√a) by both methods. Example 5: y = 1/x² at x = −1: dy/dx = −2/x³, so at −1 the gradient is 2.

The most-used rule in calculus says: bring the power down in front and reduce it by one — d(xⁿ)/dx = n xⁿ⁻¹. For positive integer n the book proves it from the binomial expansion of (x + δx)ⁿ: every term except the first two carries (δx)² or higher, so after dividing by δx and letting δx → 0 only n xⁿ⁻¹ survives. For negative n, write n = −m and the same machinery gives −m x^(−m−1), which is the same formula. 📌 Book note: the rule also holds for real n — for example d(x^(2/3))/dx = (2/3)x^(−1/3) — though the proof for rational powers is left as an exercise.

Why students love this rule: it turns differentiation into pattern-matching. x⁷ → 7x⁶; 1/x = x⁻¹ → −x⁻²; √x = x^(1/2) → (1/2)x^(−1/2). The only traps are forgetting to reduce the power (writing 7x⁷) and mishandling constants (the derivative of 3x⁴ is 12x³, not 3 × 4x⁴). Say the rule in words every time until it is automatic: “power down front, power minus one.”

The theorem of this section has a one-way arrow, and the arrow's direction is the whole lesson: if f is differentiable at a, then f is continuous at a. The proof is short: f(x) − f(a) = [(f(x) − f(a))/(x − a)] × (x − a), and as x → a the bracket tends to f′(a) while (x − a) tends to 0, so f(x) → f(a). Differentiability is stronger than continuity — it demands a smooth tangent, not just an unbroken graph.

The converse is false, and the book's counterexample is |x| at 0. The function is continuous at 0 (its graph is unbroken), but the difference quotient from the right tends to 1 and from the left to −1, so the two-sided limit does not exist: f′(0) does not exist, and there is no single tangent line at the corner. A corner kills differentiability but not continuity. Whenever a graph has a sharp point, a jump, or a vertical tangent, expect the derivative to fail there even if the function itself survives.

(1) d(c)/dx = 0 for any constant c — flat graphs, zero slope (already seen in Example 3). (2) The power rule for any real n: d(xⁿ)/dx = n xⁿ⁻¹. (3) d(cf)/dx = c f′(x): constants factor out of derivatives. The book proves it from the definition, and Example 6 applies it: d(3x^(4/3))/dx = 3 × (4/3)x^(1/3) = 4x^(1/3). (4) The sum/difference rule: d(f ± g)/dx = f′ ± g′ — differentiate term by term. 📌 Book note: the rule extends to more than two functions; a polynomial is just a sum of powers, each handled by (2) and (3).

Example 7 differentiates a polynomial term by term to get 3x³ + 2x² + x + 2 (each term handled by the power rule and the constant-multiple rule). Example 8 expands (x² + 5)(x³ + 7) first, then differentiates the expanded polynomial to get 5x⁴ + 15x² + 14x. Example 9 rewrites (2√x + 2)(x − √x) as 2(x^(3/2) − x^(1/2)) and differentiates to (3x − 1)/√x. The lesson of Examples 8 and 9: simplify before differentiating whenever you can — the product rule of the next section exists for when you cannot.

When two functions are multiplied and cannot be simplified first, use the product rule: d(uv)/dx = u dv/dx + v du/dx — “first times derivative of second, plus second times derivative of first”. The book proves it from the definition by the classic add-and-subtract trick: u(x + δx)v(x + δx) − u(x)v(x) is rewritten by inserting u(x + δx)v(x) in the middle, so the difference splits into two difference quotients, one for u and one for v.

Example 10 is a deliberate cross-check: it differentiates the same function as Example 9 — (2√x + 2)(x − √x) — but this time by the product rule instead of by expanding first. The working is longer, but the answer is the same: (3x − 1)/√x. When a book solves one question two ways and both agree, that is not repetition — it is a built-in error check you can copy in exams.

For a quotient u/v, the quotient rule is d(u/v)/dx = (v du/dx − u dv/dx)/v² — “denominator times derivative of numerator, minus numerator times derivative of denominator, all over denominator squared”. The order matters: the minus sign means swapping u and v changes the answer's sign, unlike the product rule. The book proves it from the definition with the same add-and-subtract device as the product rule.

Example 11: differentiate (2x³ − 3x² + 5)/(x² + 1). Here u = 2x³ − 3x² + 5, u′ = 6x² − 6x, v = x² + 1, v′ = 2x. So the derivative is [(x² + 1)(6x² − 6x) − (2x³ − 3x² + 5)(2x)]/(x² + 1)². Expand the top: (6x⁴ − 6x³ + 6x² − 6x) − (4x⁴ − 6x³ + 10x) = 2x⁴ + 6x² − 16x. So the answer is (2x⁴ + 6x² − 16x)/(x² + 1)². Always expand and cancel at the end — a factored numerator hides nothing, but an unexpanded one hides sign errors.

Derivatives earn their keep as rates in disguise. Example 12 (marginal profit): P(x) = 100x − 5x² gives P′(x) = 100 − 10x, so at x = 8 the marginal profit is P′(8) = 100 − 80 = 20 — roughly, one more unit sold near x = 8 adds 20 to profit. Example 13 (motion): s = 4t³ − 3t² + 2t gives v = s′ = 12t² − 6t + 2 and a = v′ = 24t − 6; at t = 2, v = 48 − 12 + 2 = 38 m/s and a = 48 − 6 = 42 m/s². Example 14 (growth): V = 5000(1 + 0.04t)² gives V′ = 5000 × 2(1 + 0.04t)(0.04) = 400(1 + 0.04t); at t = 10, V′ = 400 × 1.4 = 560 per year. Example 15 (pressure gradient): S = 500x − 2x³ gives S′ = 500 − 6x²; at x = 5, S′ = 500 − 150 = 350 Pa/m.

The pattern across all four: name the function, differentiate it, evaluate at the asked point, and attach the unit. Students lose marks not on the calculus but on the last step — a bare number without “per year” or “m/s” is an unfinished answer.