Mathematics: Calculus

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Calculus is the mathematical study of continuous change. It has two main branches: Differential Calculus (rates of change and slopes of curves) and Integral Calculus (areas under curves and accumulation of quantities).

Limits — The Foundation of Calculus: A limit describes the value a function approaches as the input approaches a specific value. Notation: $\lim_{x \to a} f(x) = L$.

Basic limit rules: $$\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$ $$\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$$ $$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$$

Standard Limits:

• $\lim_{x \to 0} \frac{\sin x}{x} = 1$ (with $x$ in radians)
• $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$
• $\lim_{x \to \infty} \frac{\sin x}{x} = 0$

Derivatives — Rate of Change: The derivative of $f(x)$ is defined as: $$f’(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \frac{dy}{dx}$$

Derivative Rules:

Function	Derivative
$x^n$	$nx^{n-1}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$e^x$	$e^x$
$\ln x$	$\frac{1}{x}$
$cf(x)$	$cf’(x)$
$f(x) + g(x)$	$f’(x) + g’(x)$
$f(x) \cdot g(x)$	$f’g + fg’$ (product rule)
$\frac{f(x)}{g(x)}$	$\frac{f’g - fg’}{g^2}$ (quotient rule)

⚡ HAT-UG Exam Tip: For the product rule, use the mnemonic “first times derivative of second plus second times derivative of first” or remember: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$. For the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u’ - u v’}{v^2}$.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding and consistent scores.

The Chain Rule — Composite Functions: If $y = f(g(x))$, then: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \quad \text{where } u = g(x)$$

Example: Find the derivative of $y = (3x^2 + 2)^5$.

• Let $u = 3x^2 + 2$, so $y = u^5$
• $\frac{dy}{du} = 5u^4 = 5(3x^2 + 2)^4$
• $\frac{du}{dx} = 6x$
• $\frac{dy}{dx} = 5(3x^2 + 2)^4 \cdot 6x = 30x(3x^2 + 2)^4$

Derivatives of Trigonometric Functions: $$\frac{d}{dx}(\tan x) = \sec^2 x$$ $$\frac{d}{dx}(\cot x) = -\csc^2 x$$ $$\frac{d}{dx}(\sec x) = \sec x \tan x$$ $$\frac{d}{dx}(\csc x) = -\csc x \cot x$$

Second Derivative — Acceleration: The second derivative $f”(x) = \frac{d^2y}{dx^2}$ tells you whether a function is concave up or down. In kinematics, if $y$ represents position and $x$ represents time, then $\frac{dy}{dx}$ is velocity and $\frac{d^2y}{dx^2}$ is acceleration.

Integration — The Reverse of Differentiation: $$\int x^n , dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$$ $$\int e^x , dx = e^x + C$$ $$\int \frac{1}{x} , dx = \ln|x| + C$$ $$\int \sin x , dx = -\cos x + C$$ $$\int \cos x , dx = \sin x + C$$

Definite Integrals: $$\int_a^b f(x) , dx = F(b) - F(a)$$ where $F(x)$ is an antiderivative of $f(x)$.

⚡ Standard Study Tip: When you differentiate, you lose the constant. When you integrate, you must add $+C$ (the constant of integration) for indefinite integrals. This constant often matters in applied problems — for example, when finding a specific curve from its gradient, the constant is determined by a boundary condition.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory with derivations, historical context, and exam patterns.

The Fundamental Theorem of Calculus:

This theorem links differentiation and integration — the two branches of calculus. It has two parts:

Part 1: If $F(x) = \int_a^x f(t) , dt$, then $F’(x) = f(x)$. In other words, the derivative of an integral is the original function.

Part 2: If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then: $$\int_a^b f(x) , dx = F(b) - F(a)$$

This theorem, independently discovered by Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century, transformed mathematics. Newton used calculus to derive the laws of planetary motion; Leibniz developed the notation ($d$, $\int$) that we still use today.

Derivative of Exponential and Logarithmic Functions:

Starting from $y = e^x$, we find $\frac{dy}{dx}$ using the definition: $$\frac{d}{dx}(e^x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h} = e^x \cdot 1 = e^x$$

For $y = \ln x$: if $y = \ln x$, then $x = e^y$. Differentiating implicitly: $\frac{dx}{dy} = e^y = x$, so $\frac{dy}{dx} = \frac{1}{x}$.

Integration by Parts — The “Reverse” of the Product Rule: $$\int u , dv = uv - \int v , du$$

Choose $u$ using the LIATE rule (in order of preference for $u$): Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.

Example: $\int x e^x , dx$

• Let $u = x$ (algebraic), $dv = e^x dx$ (exponential)
• Then $du = dx$, $v = e^x$
• $\int x e^x , dx = x e^x - \int e^x , dx = x e^x - e^x + C = e^x(x - 1) + C$

Area Under a Curve: The definite integral $\int_a^b f(x) , dx$ gives the signed area between $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$. For $f(x) > 0$, this is the actual area. For $f(x) < 0$, the integral gives a negative value.

Example: Find the area under $y = x^2$ from $x = 0$ to $x = 3$. $$\int_0^3 x^2 , dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9 \text{ square units}$$

HAT-UG Calculus — Past Year Patterns (2019–2024):

• Limits: 1–2 questions (including L’Hôpital’s rule for higher-tier students)
• Derivatives: 2–3 questions (including chain rule applications)
• Second derivatives and increasing/decreasing functions: 1 question
• Integration: 2–3 questions (definite integrals, area)
• Integration by parts: appears in 1 in 4 papers
• Exponential and log derivatives: 1 question

⚡ HAT-UG Advanced Strategy: If a limit takes the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, apply L’Hôpital’s Rule: differentiate the numerator and denominator separately and take the limit again. Example: $\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1$. For repeated applications, differentiate again and again until the limit resolves.

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