12.2 Calculus Basics

12.2 Calculus Basics

Differentiation

Differentiation is the process of finding the derivative — the instantaneous rate of change of a function. Geometrically, the derivative dy/dx at a point is the slope of the tangent to the curve y = f(x) at that point.

The power rule is the workhorse of FSc calculus:

• d/dx (xⁿ) = n·xⁿ⁻¹ (for any real n)

Combined with the constant multiple rule, d/dx (a·xⁿ) = a·n·xⁿ⁻¹, and the sum rule, d/dx (f + g) = f' + g', you can differentiate any polynomial term by term.

Worked example: Find dy/dx for y = 5x³ − 4x² + 2x − 7.

• dy/dx = 15x² − 8x + 2

Apply the power rule term by term: the constant −7 vanishes (derivative of a constant is 0).

Worked example — fractional exponent: y = 4√x = 4x^(1/2).

• dy/dx = 4 · (1/2) · x^(−1/2) = 2 / √x

Product, Quotient, and Chain Rules

For functions built from products, quotients, or compositions, three rules extend the power rule:

• Product rule: (uv)' = u'v + uv'
• Quotient rule: (u/v)' = (u'v − uv') / v²
• Chain rule: d/dx f(g(x)) = f'(g(x)) · g'(x)

Worked example — product rule: y = x² · sin x. Here u = x², v = sin x, so u' = 2x and v' = cos x.

• dy/dx = 2x · sin x + x² · cos x

Worked example — quotient rule: y = (x² + 1) / x. Here u = x² + 1, v = x, so u' = 2x and v' = 1.

• dy/dx = (2x · x − (x² + 1) · 1) / x² = (2x² − x² − 1) / x² = (x² − 1) / x²

Worked example — chain rule: y = (3x + 1)⁵.

• dy/dx = 5(3x + 1)⁴ · 3 = 15(3x + 1)⁴

The chain rule is essential whenever a function is "nested" inside another — recognising the inner function (g = 3x + 1 here) is the key step.

Table of Derivative and Integral Rules

Memorise this table — the GD Pilot test rarely goes beyond these standard forms. Note the symmetry: the derivative of eˣ is itself, and the integral of 1/x is ln|x|, the one case where the power rule breaks down (n = −1).

Integration and Definite Integrals

Integration is the reverse of differentiation — it finds the antiderivative. The indefinite integral of xⁿ (n ≠ −1) is:

• ∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C

The constant C appears because differentiation "loses" constants. A definite integral ∫ from a to b gives a number — geometrically, the signed area under the curve between x = a and x = b.

Worked example — definite integral: ∫₀² (3x² + 2) dx.

• Antiderivative: x³ + 2x
• Evaluate: [(2)³ + 2(2)] − [(0)³ + 2(0)] = 8 + 4 − 0 = 12

So the area under y = 3x² + 2 from x = 0 to x = 2 is 12 square units.

Applications

Calculus is not abstract on the GD Pilot test — it shows up in concrete problems:

• Rate of change: If s(t) is position, then v = ds/dt is velocity and a = dv/dt = d²s/dt² is acceleration.
• Area under a curve: ∫ from a to b of f(x) dx gives the geometric area (signed).
• Maxima and minima: At a turning point, dy/dx = 0. Use the second derivative: if d²y/dx² < 0, it's a maximum; if > 0, a minimum.

Worked example — optimisation: A rectangular field is fenced along a river, so only three sides need fencing. If 60 m of fence is available, find the dimensions that maximise area. Let the two sides perpendicular to the river be x each, and the parallel side be y. Then 2x + y = 60, so y = 60 − 2x. Area A = x·y = x(60 − 2x) = 60x − 2x².

• dA/dx = 60 − 4x = 0 ⇒ x = 15
• y = 60 − 2(15) = 30
• Maximum area = 15 × 30 = 450 m²

Second derivative d²A/dx² = −4 < 0, confirming a maximum.

Extra Differentiation Practice

Worked example — trig with the chain rule: Find dy/dx for y = sin(5x² + 1).

• Outer function f(u) = sin u ⇒ f'(u) = cos u
• Inner function g(x) = 5x² + 1 ⇒ g'(x) = 10x
• dy/dx = cos(5x² + 1) · 10x = 10x · cos(5x² + 1)

A reliable check: the derivative of a composite always keeps the inner function unchanged inside the outer — only the outer gets differentiated, and the inner derivative is multiplied on the outside. Forgetting the inner derivative (the trailing 10x here) is the single most common chain-rule error on the test.

Worked example — second derivative test: For y = x³ − 3x² + 2, locate and classify the turning points.

• dy/dx = 3x² − 6x = 3x(x − 2) = 0 ⇒ x = 0 or x = 2
• d²y/dx² = 6x − 6
• At x = 0: d²y/dx² = −6 < 0 ⇒ local maximum at (0, 2)
• At x = 2: d²y/dx² = 6 > 0 ⇒ local minimum at (2, −2)

If the second derivative is exactly zero, the test is inconclusive and you must inspect the sign change of dy/dx around the point.

Area Between Two Curves

The definite integral extends naturally to the area between two curves. If f(x) ≥ g(x) on [a, b], the enclosed area is:

• A = ∫ from a to b of [f(x) − g(x)] dx

Worked example: Find the area enclosed between y = x + 2 and y = x² on 0 ≤ x ≤ 2.

• On [0, 2], x + 2 ≥ x² (check at x = 1: 3 ≥ 1 ✓), so the line is the upper curve.
• A = ∫₀² [(x + 2) − x²] dx = [x²/2 + 2x − x³/3]₀²
• Evaluate: [(4/2 + 4 − 8/3)] − [0] = 2 + 4 − 8/3 = 6 − 8/3 = (18 − 8)/3 = 10/3 square units

This is the same machinery as a single-curve definite integral — the integrand is simply the difference of the two functions. Always confirm which curve is on top before integrating; if the curves cross within the interval, split the integral at each crossing point and flip the order so each sub-integral stays non-negative.