Calculus from the class notes, with pictures

This page covers everything in the class notes (functions → linear systems) and all 7 questions of Assignment 1, due Wed 14 Oct 2026. Every section gives you the idea, a picture you can play with, the formulas to remember, and a short "how to teach it" box for explaining it to your classmate.

How the topics build on each other

Limits are the foundation. Continuity is defined with a limit, the derivative is a limit (of secant slopes), and the definite integral is a limit (of Riemann sums). Linear systems are the start of the next unit (Linear Algebra).

Which section each assignment question needs

Functions & combining them

A function is a rule that gives each input exactly one output. The domain is the set of inputs the rule accepts. In this course, only three things remove an input from the domain:

1. Division by zero: $\frac{1}{x-1}$ rejects $x=1$.
2. An even root of a negative number: $\sqrt{x}$ needs $x\ge 0$.
3. The log of a number $\le 0$: $\ln x$ needs $x>0$ (this shows up in Q2).

The example from the notes uses $f(x)=\sqrt x$ and $g(x)=\sqrt{1-x}$:

Composite functions

$(f\circ g)(x)=f(g(x))$ means you apply $g$ first, then put its output into $f$. Think of two machines connected in a line:

The example from the notes uses $f(x)=\sqrt x$ (domain $[0,\infty)$) and $g(x)=x+1$ (domain $\mathbb R$). The notes leave $f\circ f$ and $g\circ g$ blank, so they're filled in here:

Order matters: $\sqrt{x+1}\ne\sqrt x+1$. In general $f\circ g\neq g\circ f$.

Even, odd and piecewise functions

How to test: replace $x$ with $-x$ and simplify. If you get back $f(x)$, it's even. If you get $-f(x)$, it's odd. If you get neither, it's neither (most functions are, e.g. $x^2+x$).

Piecewise function: uses a different rule on different parts of the domain. The notes' example is $$|x|=\begin{cases}x,&x\ge 0\\-x,&x<0\end{cases}$$ Always check which piece owns the boundary point ($\le$ vs $<$). Q2 and Q5 both depend on it.

Limits

$\displaystyle\lim_{x\to c}f(x)=L$ means that as $x$ gets close to $c$ (without ever being equal to $c$), $f(x)$ gets close to $L$. The value of $f(c)$ itself doesn't matter, and $f(c)$ doesn't even need to exist.

The example from the notes: $\displaystyle\lim_{x\to0}\frac{\sin 4x}{\sin 2x}=2$

Plugging in $x=0$ gives $\tfrac00$, which tells you nothing yet. Use the standard limit $\displaystyle\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$:

Alternatively, since $\sin4x=2\sin2x\cos2x$, the ratio equals $2\cos2x\to2$.

Toolkit for computing limits

1. Direct substitution. If plugging in gives a normal number, that's the answer.
2. Factor and cancel when you get $\tfrac00$.
3. Multiply by the conjugate when you see a difference of square roots: $(\sqrt A-\sqrt B)\cdot\frac{\sqrt A+\sqrt B}{\sqrt A+\sqrt B}=\frac{A-B}{\sqrt A+\sqrt B}$. The notes use this to differentiate $\sqrt x$ (§7), and Q3 needs it.
4. The standard limit $\frac{\sin\theta}{\theta}\to1$.

Continuity

Informally, a function is continuous if you can draw its graph without lifting the pencil. The formal test at a point $x=c$ has three parts:

On a closed interval $[a,b]$: $f$ has to be continuous at every interior point $c$, plus be one-sided continuous at the two ends:

Intermediate Value Theorem (IVT)

Why continuity is required: a function with a jump can skip straight over $N$. Look at the jump at $x=1$ in the §5 demo: the graph never takes the value 2 near there. Also note that IVT only tells you a root exists. It doesn't tell you where. Bisection (in the demo) is one way to locate it.

Differentiability & derivative rules

• $f$ is differentiable on an open interval if it has a derivative at every point of the interval.
• $f$ is differentiable on a closed interval $[a,b]$ if, in addition, the one-sided derivatives exist at the endpoints $a$ and $b$.

Example 1 (notes): $f(x)=\sqrt x$ by first principle

1. $f'(x)=\displaystyle\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h}$ gives $\tfrac00$ at $h=0$.
2. Multiply by the conjugate: $\displaystyle\frac{\sqrt{x+h}-\sqrt x}{h}\cdot\frac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x}=\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt x)}=\frac{1}{\sqrt{x+h}+\sqrt x}$
3. Now substitute $h=0$: $f'(x)=\dfrac{1}{2\sqrt x}$ (for $x>0$).

Example 2 (notes): $f(x)=|x|$ at $x=0$

LHD ≠ RHD, so $|x|$ is not differentiable at 0. The graph has a sharp corner there, so no single tangent line fits.

Rules (u = u(x), v = v(x))

Product rule: proof from first principle (notes Q1)

1. $\dfrac{d}{dx}(uv)=\displaystyle\lim_{h\to0}\frac{u(x+h)v(x+h)-u(x)v(x)}{h}$
2. Add and subtract $u(x+h)\,v(x)$ in the numerator. This adds zero, but it lets you factor: $$=\lim_{h\to0}\frac{u(x+h)\,[v(x+h)-v(x)]+v(x)\,[u(x+h)-u(x)]}{h}$$
3. Split the limit: $\displaystyle\lim u(x+h)\cdot\frac{v(x+h)-v(x)}{h}+\lim v(x)\cdot\frac{u(x+h)-u(x)}{h}$
4. As $h\to0$: $u(x+h)\to u(x)$ (because $u$ is differentiable, hence continuous), and the two fractions become $\frac{dv}{dx}$ and $\frac{du}{dx}$. So $\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ ∎

Quotient rule: the homework proof

1. $\dfrac{d}{dx}\Big(\dfrac uv\Big)=\displaystyle\lim_{h\to0}\frac1h\left[\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}\right]=\lim_{h\to0}\frac{u(x+h)v(x)-u(x)v(x+h)}{h\,v(x+h)\,v(x)}$
2. Add and subtract $u(x)v(x)$ in the numerator: $$u(x+h)v(x)-u(x)v(x)\;-\;u(x)v(x+h)+u(x)v(x)=v(x)\,[u(x+h)-u(x)]-u(x)\,[v(x+h)-v(x)]$$
3. Divide by $h$ and let $h\to0$: the numerator becomes $v\frac{du}{dx}-u\frac{dv}{dx}$, and the denominator $v(x+h)v(x)\to v^2$. $$\frac{d}{dx}\Big(\frac uv\Big)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\quad(v\ne0)\ \ ∎$$

Higher-order, total and partial derivatives

• 1st order: $\frac{df}{dx}$. 2nd order: $f''(x)=\frac{d}{dx}\big(\frac{df}{dx}\big)$. nth order: $\frac{d^nf}{dx^n}$ means differentiate $n$ times. For example, if $f=x^3$, then $f'=3x^2$, $f''=6x$, $f'''=6$.
• Total derivative: when $f$ depends on one variable, $f'(x)=\frac{df(x)}{dx}$.
• Partial derivative: when $f$ depends on several variables, like $f(x,y)$, $\frac{\partial f}{\partial x}$ differentiates in $x$ while treating $y$ as a constant. For example, with $f=x^2y+3y$: $\frac{\partial f}{\partial x}=2xy$ and $\frac{\partial f}{\partial y}=x^2+3$.

Antiderivatives & the indefinite integral

From the notes: $\int 2x\,dx=x^2+C$, because $\frac{d}{dx}(x^2+5)=2x$ and $\frac{d}{dx}(x^2+8)=2x$. The constant disappears when you differentiate, so you can't tell which one you started from, and you write $+C$.

The definite integral

$\displaystyle\int_a^b f(x)\,dx$ is the integral over a closed interval $[a,b]$. It equals the (signed) area under the curve. Area above the x-axis counts as +, and area below counts as −.

Mean value theorem for definite integrals

The worked example from the notes

$f(x)=\begin{cases}x+4,&-4\le x\le-1\\-x+2,&-1<x\le2\end{cases}$ on $[-4,2]$. Find the average value.

1. $f(c)=\dfrac{1}{2-(-4)}\left[\displaystyle\int_{-4}^{-1}(x+4)\,dx+\int_{-1}^{2}(-x+2)\,dx\right]$ (splitting at the joint uses property 3)
2. $\displaystyle\int_{-4}^{-1}(x+4)dx=\Big[\tfrac{x^2}{2}+4x\Big]_{-4}^{-1}=\big(\tfrac12-4\big)-\big(8-16\big)=-3.5+8=4.5$
3. $\displaystyle\int_{-1}^{2}(-x+2)dx=\Big[-\tfrac{x^2}{2}+2x\Big]_{-1}^{2}=\big(-2+4\big)-\big(-\tfrac12-2\big)=2+2.5=4.5$
4. $f(c)=\dfrac{1}{6}(4.5+4.5)=\dfrac96=\dfrac32$ average value = 3/2
5. Bonus: where does $f(c)=\tfrac32$? $x+4=1.5\Rightarrow c=-2.5$, and $-x+2=1.5\Rightarrow c=0.5$. There are two such points, both inside $[-4,2]$.

Systems of linear equations

The example from the notes:

1. From (2): $y=5-x$.
2. Substitute into (1): $2x-(5-x)=1\Rightarrow 3x-5=1\Rightarrow x=2$.
3. $y=5-2=3$. (x, y) = (2, 3)

Geometric picture 1: the row picture

Each row (each equation) is a line. The solution is the point where the lines intersect. There are only three possibilities: one point (unique solution), parallel lines (no solution), or the same line twice (infinitely many solutions).

Geometric picture 2: the column picture

Read the same system by columns. You're asking which combination of the column vectors produces the right-hand side:

The 3 × 3 system in the notes

In the row picture, each equation is a plane in 3-D, and the solution is the point where all three planes meet. The notes don't solve this one, so here's the elimination:

1. (1) − (2): $x-2y=-1\Rightarrow x=2y-1$
2. Put that into (2): $(2y-1)+y+z=2\Rightarrow z=3-3y$
3. Put both into (3): $4(2y-1)+6y+9(3-3y)=10\Rightarrow -13y+23=10\Rightarrow y=1$
4. $x=1,\ z=0$. (1, 1, 0) Check: $2-1+0=1$ ✓, $1+1+0=2$ ✓, $4+6+0=10$ ✓

In matrix form this is $A\mathbf x=\mathbf b$, with $A=\begin{bmatrix}2&-1&1\\1&1&1\\4&6&9\end{bmatrix}$, which is where the Linear Algebra unit starts.

All 7 questions, solved

Try each question first. Open the hint only if you're stuck, then compare your work with the full write-up. The write-ups are laid out the way you'd write them in the answer sheet.

A 45-minute lesson plan + quick quiz

Quick quiz (answers hidden)

Corrections to the handwritten notes

• "F(x) is derivative of f(x)" has it backwards. $F$ is the antiderivative, and $f$ is the derivative of $F$: $F'(x)=f(x)$.
• Even function: one page writes $\int_{-a}^a f\,dx=\int_0^a f\,dx$. It should be $2\int_0^a f\,dx$ (another page has it right).
• Odd function: one page writes $=\int_{-a}^0 f\,dx$ as the result. The correct result is $\int_{-a}^a f\,dx=0$.
• $|x|$ at 0: on one page the LHD reads like $+1$. It is $-1$ ($\frac{-h}{h}$). That difference is exactly why $|x|$ is not differentiable at 0.
• Power-of-a-function rule: written as "$n\frac{d}{dx}f(x)^{n-1}$". The correct form is $\frac{d}{dx}[f(x)]^n=n[f(x)]^{n-1}\cdot f'(x)$.
• $3g$ row of the domain table: $3g=3\sqrt{1-x}$, so its domain is $(-\infty,1]$, the same as $g$.
• Product-rule proof, first line on one page: "$u(x+\Delta x)-v(x-\Delta x)$" is a slip. The rest of that proof (add and subtract $u(x+h)v(x)$) is correct. Ignore the crossed-out attempt on the other page.
• $\int(u\cdot v)\,dx$ is left blank, which is fine. It is not $\int u\cdot\int v$ (see the §8 trap).