Calculus and Differential Equations: Derivatives, Optimization, Integrals, ODEs, and Laplace Basics

Calculus questions are graded purely on whether your final number is right, and nearly every item reduces to a short list of rules applied in the right order. This lesson works through limits, differentiation, optimization, integration for area and volume, first-order differential equations, and the handful of Laplace transforms you carry into circuits later, the exact sequence the board tests.

Limits and derivatives

A limit describes the value a function approaches as x approaches some point, even if the function is undefined exactly there. When direct substitution gives 0/0, factor and cancel first.

Worked example: Find the limit as x approaches 3 of (x^2 - 9)/(x - 3). Factor: (x-3)(x+3)/(x-3) = x + 3 for x != 3, so the limit is 3 + 3 = 6. A second common form uses the identity sin(kx)/x -> k as x -> 0: the limit as x approaches 0 of sin(5x)/x is 5.

Rule	Formula
Power rule	d/dx[x^n] = n x^(n-1)
Product rule	d/dx[uv] = u'v + uv'
Quotient rule	d/dx[u/v] = (u'v - uv') / v^2
Chain rule	d/dx[f(g(x))] = f'(g(x)) * g'(x)

Worked example (power rule): f(x) = 5x^4 - 3x^3 + 2x differentiates term by term to f'(x) = 20x^3 - 9x^2 + 2. Worked example (product rule): f(x) = x^3 ln(x) gives f'(x) = 3x^2 ln(x) + x^3(1/x) = 3x^2 ln(x) + x^2. Worked example (chain rule): f(x) = sqrt(4x^2 + 9) = (4x^2+9)^(1/2) gives f'(x) = (1/2)(4x^2+9)^(-1/2)(8x) = 4x / sqrt(4x^2+9).

Maxima, minima, and optimization

A critical point occurs where f'(x) = 0. The second-derivative test classifies it: f''(x) > 0 is a relative minimum, f''(x) < 0 is a relative maximum. Optimization word problems follow a fixed routine: express the target quantity as a function of one variable using the given constraint, differentiate, set the derivative to zero, and check the second derivative or the domain endpoints.

Worked example: For f(x) = x^3 - 3x^2 - 9x + 5, f'(x) = 3x^2 - 6x - 9 = 3(x-3)(x+1), so the critical points are x = 3 and x = -1. Since f''(x) = 6x - 6, f''(-1) = -12 < 0 (a relative maximum) and f''(3) = 12 > 0 (a relative minimum). At x = -1, f(-1) = -1 - 3 + 9 + 5 = 10. Worked example (optimization): An open-top box is formed by cutting equal squares of side x from the corners of a 30 cm by 30 cm sheet and folding up the sides. Volume V(x) = x(30 - 2x)^2. Setting dV/dx = (30-2x)(30-6x) = 0 gives x = 15 (a boundary point where V = 0) or x = 5 (the interior maximum), and V(5) = 5(20)^2 = 2000 cm^3.

Integral calculus: area and volume

The definite integral of f(x) from a to b gives the (signed) area under the curve. When a region bounded by a curve and the x-axis is revolved about the x-axis, the disk method gives its volume: V = pi * integral from a to b of [f(x)]^2 dx.

Worked example (definite integral): The integral of (3x^2 - 4x + 1) dx from x = 2 to x = 4 uses the antiderivative x^3 - 2x^2 + x. Evaluating: F(4) = 64 - 32 + 4 = 36, F(2) = 8 - 8 + 2 = 2, so the integral is 36 - 2 = 34. Worked example (area between curves): y = x^2 and y = 2x intersect at x = 0 and x = 2, with y = 2x on top in between. The area is the integral of (2x - x^2) dx from 0 to 2 = [x^2 - x^3/3] at 2 = 4 - 8/3 = 4/3. Worked example (volume of revolution): Revolve the region under y = 3x from x = 0 to x = 2 about the x-axis: V = pi * integral from 0 to 2 of (3x)^2 dx = 9 pi [x^3/3] from 0 to 2 = 9 pi(8/3) = 24 pi. As a check, this is a cone of radius 6 and height 2, and the cone formula (1/3) pi r^2 h = (1/3) pi(36)(2) = 24 pi matches exactly.

First-order differential equations

A separable equation can be rearranged so all y-terms sit with dy and all x-terms sit with dx, then both sides are integrated.

Worked example: Solve dy/dx = x/y given y = 3 when x = 0. Separate: y dy = x dx. Integrating both sides gives y^2/2 = x^2/2 + C, so y^2 = x^2 + C'. Using y(0) = 3: C' = 9, so y^2 = x^2 + 9. At x = 4, y^2 = 16 + 9 = 25, so y = 5 (the positive root). The same separable pattern models growth and decay: for dP/dt = -kP with P(0) = 1000 g and a half-life of 4 hours, e^(-4k) = 0.5, so the mass at t = 8 hours is P(8) = 1000(e^(-4k))^2 = 1000(0.5)^2 = 250 g.

Laplace transforms: the basics

The Laplace transform turns a differential equation in t into an algebraic equation in s, which is exactly why it drives transient circuit analysis later. Memorize the handful of transforms below; every table lists them.

f(t)	L{f(t)} = F(s)
1	1/s
e^(at)	1/(s - a)
t	1/s^2
sin(wt)	w/(s^2 + w^2)

Worked example: L{1} = integral from 0 to infinity of e^(-st) dt = 1/s, for s > 0. Going the other direction (inverse Laplace), since L{e^(-at)} = 1/(s+a), the inverse Laplace transform of F(s) = 5/(s + 3) is 5 e^(-3t).

A word on probability and engineering economy

Two more math-group topics ride along with calculus on exam day: probability and statistics (favorable outcomes over total outcomes, combinations for unordered selections, mean and standard deviation for describing data) and engineering economy (simple interest F = P(1 + i*n), compound interest F = P(1 + i)^n, and present worth P = F/(1 + i)^n). Both are computation-heavy in the same way calculus is: identify the right formula, plug in carefully, and recompute before you commit to an answer.

Exam-day strategy

• Factor before you substitute on any 0/0 limit; canceling the common factor is almost always the whole trick.
• Differentiate a composite function from the outside in: identify the outer function first, then multiply by the derivative of what is inside it.
• On optimization problems, reduce to a function of a single variable before differentiating; a leftover second variable means you skipped the constraint equation.
• Volume-of-revolution answers commonly carry pi; leave pi in the answer unless the item explicitly asks for a decimal approximation.
• For a differential equation, apply the initial condition last, after integrating, to solve for the constant C.
• Keep a mental Laplace table for 1, e^(at), t, and sin(wt); most Laplace items on the board are a direct lookup, not a derivation.