Shunya / Calculus Lab

Shunya · Maths lab

Calculus lab

Type any function and watch calculus happen. Squeeze x towards a point to see a limit, shrink a secant into a tangent to get the derivative, find where a curve peaks and dips, solve real optimisation problems, watch rectangles add up to an integral, and follow a differential equation through its slope field.

Lessons

Pick your class, follow the steps in the lab, and check what you see against the working under the graph.

Class 11 — Limits

Concept: the limit of f(x) as x → a is the value f(x) is heading for, from both sides, even if f(a) is missing. In symbols, $\lim_{x \to a} f(x) = L$ .

Open Limits and press (x² − 1) ÷ (x − 1) at 1. The graph has a hole at x = 1, yet the table shows both sides heading to 2.
Drag the green point towards a from the left, then from the right, and watch f(x) settle.
Press |x| ÷ x at 0: the left limit is −1 and the right limit is +1, so the limit does not exist (a jump).
Try 1 ÷ x² (both sides run to +∞) and sin(1/x) (it never settles).
Press sin x ÷ x at 0 and read the famous limit 1. Then type your own function in the box.

Takeaway: a limit exists only when the left and right limits agree on a number.

Class 11 — The derivative as a slope

Concept: the derivative is the limit of the slope of a secant as its two points come together: $f'(x_0) = \lim_{h \to 0} \dfrac{f(x_0 + h) - f(x_0)}{h}$ .

Open Derivative with f(x) = x² and x₀ = 1. The orange secant joins x₀ and x₀ + h.
Press Shrink h towards 0: the secant turns into the teal tangent, and the secant slope approaches f′(1) = 2.
Read the table of quotients under the graph: 3.5, 3, 2.1, 2.01, 2.001 … → 2.
Tick show the derivative graph and drag x₀: the blue point is the slope at each place.
Try sin x, eˣ and ln x, and compare with the symbolic derivative shown at the top.

Takeaway: the derivative is the slope of the tangent, found by letting the second point of the secant slide into the first.

Class 12 — Continuity and differentiability

Concept: f is continuous at a if f(a) exists, the limit exists and they are equal. It is differentiable at a if the slopes from the left and the right agree. Differentiable always means continuous, but not the other way round.

Open Limits and press x² at 2: continuous, and both slopes agree, so differentiable.
Press |x| at 0: continuous (no break) but the slopes are −1 and +1, so not differentiable: a corner.
Press floor x at 1: a jump, so not continuous, and therefore not differentiable either.
Press (x² − 1) ÷ (x − 1) at 1: a removable discontinuity, because the limit exists but f(1) does not.
Open Derivative and choose |x| or x^(1/3): see a corner and a vertical tangent.

Takeaway: corners, jumps and vertical tangents are the three ways a derivative can fail.

Class 12 — Increasing, decreasing, maxima and minima

Concept: where f′ > 0 the function increases and where f′ < 0 it decreases. A local maximum is where f′ changes from + to −, and a local minimum from − to +. The second derivative test uses the sign of f″.

Open Maxima & minima with x³ − 3x. The curve is green where it rises and orange where it falls.
Read the working: f′ = 3x² − 3 = 0 at x = ±1. At x = −1, f″ = −6 < 0 (a local maximum); at x = 1, f″ = 6 > 0 (a local minimum).
Drag the green point and watch the sign of f′ and f″. Tick the concavity shading to see where the curve bends up or down.
Press x³ (no max or min): f′ = 0 at x = 0 but does not change sign, so it is a horizontal inflection.
Drag the interval dots to see the absolute maximum and minimum on a closed interval.

Takeaway: stationary points are where f′ = 0, and f′ or f″ tells you what each one is.

Class 12 — Optimisation

Concept: to find the best value of a quantity, write it as a function of one variable, differentiate, set the derivative to zero, and check with the second derivative.

Open Optimisation with the rectangle of fixed perimeter. Drag the point along the area graph: it peaks at x = P ÷ 4, a square.
Read the four steps in the working: A(x) = x(P/2 − x), A′(x) = P/2 − 2x, A′ = 0 → x = P ÷ 4, and A″ = −2 < 0.
Change the problem to an open box from a square sheet and check that the best cut is s ÷ 6.
Try the can of fixed volume: the least metal is used when the height equals the diameter.
Move the slider for the perimeter, sheet size or volume and see the best value change with it.

Takeaway: the maximum or minimum sits where the slope of the graph is zero.

Class 12 — Area under a curve and the definite integral

Concept: the area under a curve is approximated by thin rectangles, $\sum f(x_i)\,\Delta x$ , and the integral $\int_a^b f(x)\,dx$ is the limit as the strips become infinitely thin.

Open Integrals with x² from 0 to 2 and the Left sum with n = 8. Read the sum and the exact value 8/3 = 2.6667.
Press n = 20 then n = 100: the error shrinks in proportion to Δx.
Switch to Midpoint and Trapezoid: they are far more accurate with the same n.
Tick show the area function: its slope at b equals f(b). This is the fundamental theorem of calculus.
Press x³ − 2x (negative parts) and compare the signed integral with the total area.

Takeaway: the integral is a limit of sums, and it is computed as F(b) − F(a) where F′ = f.

Class 12 — Differential equations

Concept: a differential equation relates a function to its derivative. Its solution is a curve whose slope at every point matches the equation, and the initial point picks out one solution.

Open Differential equations with dy/dx = x + y. Every short line is the slope at that point.
Drag the green start point: the blue curve follows the slope field, and the working gives the general solution y = Ce^x − x − 1.
Tick Euler's method (orange). Lower the Euler step h and watch it close in on the blue curve.
Press dy/dx = y(1 − y), the logistic curve, and dy/dx = −x ÷ y, whose solutions are circles.
Compare the exact, Runge–Kutta and Euler values on the cards.

Takeaway: a first-order equation gives a slope at every point; separating the variables or using the slope field finds the curve.

Class 12 — Limits at infinity, L'Hôpital's rule and the squeeze theorem

Concept: a limit at infinity asks what f(x) settles to when x is huge. L'Hôpital's rule turns a 0 ÷ 0 or ∞ ÷ ∞ limit into the limit of (top′ ÷ bottom′). The squeeze theorem traps a wobbly function between two simple ones with the same limit.

Open Limits and press (2x² + 3x) ÷ (x² + 1) at +∞. The values 2.2772, 2.0298, 2.003 (at x = 10, 100, 1000) settle on 2, so y = 2 is a horizontal asymptote. The working also shows L'Hôpital: (4x + 3) ÷ 2x, then 4 ÷ 2 = 2.
(1 + 1/x)ˣ at +∞: 2.5937, 2.7048, 2.7169 … → e = 2.7183. This is a 1^∞ form, not a quotient, so L'Hôpital does not apply: that is a common trap.
(x − sin x) ÷ x³ at 0 is 0 ÷ 0. The working differentiates three times: (1 − cos x) ÷ 3x², then sin x ÷ 6x, then cos x ÷ 6 → 1/6. For (tan x − x) ÷ x³ the same three rounds give 1/3.
eˣ ÷ x² at +∞ is ∞ ÷ ∞; after two rounds it becomes eˣ ÷ 2 → +∞: exponentials beat powers. sin x ÷ x at +∞ is different: the top has no limit, so L'Hôpital is not allowed.
The squeeze: press sin x ÷ x at +∞ (squeeze). With −1/x ≤ sin x ÷ x ≤ 1/x and both bounds → 0, the limit is 0. Press x² sin(1/x) at 0 (squeeze): −x² ≤ f ≤ x², so f → 0 even though sin(1/x) itself has no limit.

Takeaway: check the form first (0 ÷ 0 or ∞ ÷ ∞), differentiate top and bottom separately, and use the squeeze when a bounded wobble is multiplied by something that → 0.

Class 12 — Rolle's theorem and the mean value theorem

Concept: if f is continuous on [a, b] and differentiable on (a, b), some tangent is parallel to the chord: f′(c) = (f(b) − f(a)) ÷ (b − a). Rolle's theorem is the case f(a) = f(b), where f′(c) = 0.

Open Rolle & Mean value theorem with x² on [0, 2]: the chord slope is (4 − 0) ÷ 2 = 2, and f′(c) = 2c = 2 gives c = 1.
x³ − 3x on [−√3, √3] has f(a) = f(b) = 0 (Rolle), and f′(c) = 3c² − 3 = 0 gives c = −1 and c = 1: two points.
|x| on [−1, 1] has equal ends but a corner at 0: it is not differentiable there, and no tangent is flat. 1 ÷ x on [−1, 1] is not even continuous. In both cases the theorem promises nothing.
x^(1/3) on [−1, 1] has a vertical tangent at 0, so the theorem does not apply, yet c = ±0.192 does exist: the conditions are sufficient, not necessary.

Takeaway: always check both conditions before using the theorem; if either fails there may or may not be a point c.

Class 12 — Area between curves and volumes of revolution

Concept: the area between y = f(x) and y = g(x) is the integral of (upper − lower); split it wherever the curves cross. Spinning the region gives a solid whose volume is a sum of thin discs (washers) or shells.

Open Area & volume with y = x and y = x² from 0 to 1: area = ∫(x − x²) dx = 1/2 − 1/3 = 1/6 = 0.1667.
x³ and x from −1 to 1: they cross at x = 0, so the signed integral is 0 but the true area is 2 × 1/4 = 0.5. Splitting at the crossing is essential.
sin x and cos x from π/4 to 5π/4: area = [−cos x − sin x] = 2√2 = 2.8284.
Turn on around the x-axis for y = √x and y = x on [0, 1]: V = π∫(x − x²) dx = π/6 = 0.5236. Around the y-axis (shells) for y = x and y = x²: V = 2π∫ x(x − x²) dx = π/6 = 0.5236 too.

Takeaway: A = ∫|f − g| dx; V = π∫(R² − r²) dx about the x-axis; V = 2π∫|x|(f − g) dx about the y-axis.

Class 12 — Integration by substitution

Concept: substitution reverses the chain rule. When the integrand contains a function g(x) and (a multiple of) its derivative g′(x), put u = g(x), replace g′(x) dx by du, and integrate in u.

Open Integration methods → Substitution. The first problem is ∫ x (x² + 2)³ dx. The derivative of the bracket is 2x, and x is outside, so u = x² + 2 and x dx = du ÷ 2. Then ∫ u³ du ÷ 2 = u⁴ ÷ 8, so the answer is (x² + 2)⁴ ÷ 8 + C.
Change the limits with u. From x = 0 to 1.5 the variable u = x² + 2 runs from 2 to 4.25, so the definite integral is (4.25⁴ − 2⁴) ÷ 8 = 38.78, and the lab confirms it against numerical integration. Never put the old limits on the new variable.
Choose x e^(kx²): with k = 1 the answer is e^(x²) ÷ 2, and from 0 to 1 the value is (e − 1) ÷ 2 = 0.8591. Choose (ln x)ⁿ ÷ x with n = 2: u = ln x gives (ln x)³ ÷ 3.
f′ ÷ f: ∫ (2x + b) ÷ (x² + bx + c) dx = ln(x² + bx + c) + C, because the top is exactly the derivative of the bottom. Also ∫ tan x dx = −ln|cos x| + C.
Trigonometric substitution: for ∫ dx ÷ √(4 − x²) put x = 2 sin θ; the radical becomes 2 cos θ and the integral is θ = sin⁻¹(x ÷ 2). From −1 to 1 it equals 2 sin⁻¹(1/2) = π/3 = 1.0472.
Open Check your answer and try ∫ x e^(x²) dx with the wrong answer e^(x²): the lab differentiates it, finds G′ = 2 × f and tells you to divide by 2. Forgetting the factor from du is the most common mistake.

Takeaway: look for a function and its derivative inside the integrand; set u, convert dx, integrate, return to x, and always differentiate to check.

Class 12 — Integration by parts and the LIATE rule

Concept: by parts reverses the product rule: ∫ u dv = uv − ∫ v du. Pick u so that du is simpler, and dv so that you can integrate it. LIATE (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) ranks the choice of u: the earlier in the list, the better u is.

Open By parts: ∫ x e^(2x) dx. x is Algebraic and e^(2x) is Exponential, so u = x, dv = e^(2x) dx, du = dx, v = e^(2x) ÷ 2. Then x e^(2x) ÷ 2 − ∫ e^(2x) ÷ 2 dx = e^(2x)(x ÷ 2 − 1 ÷ 4) + C. From 0 to 1 the value is 2.0973.
∫ ln x dx has only one visible factor: write it as 1 · ln x with u = ln x, dv = dx. The answer is x ln x − x + C, and from 1 to 4 it is 4 ln 4 − 4 + 1 = 2.5452.
x² e^x needs parts twice: the first round leaves ∫ x e^x dx, which takes a second round. x sin 2x from 0 to 2 gives −cos 4 + (sin 4) ÷ 4 = 0.4644.
The cyclic case: in ∫ e^(ax) sin(bx) dx both factors survive differentiation. After two rounds the original integral I reappears on the right, and you solve for it: I = e^(ax)(a sin bx − b cos bx) ÷ (a² + b²). Press e^(ax) sin(bx) to see all the steps.
Inverse trig functions are done by parts with dv = dx: ∫ tan⁻¹x dx = x tan⁻¹x − ½ ln(1 + x²) + C and ∫ sin⁻¹x dx = x sin⁻¹x + √(1 − x²) + C.
The eˣ shortcut: ∫ eˣ [f(x) + f′(x)] dx = eˣ f(x) + C. For f = sin x it gives eˣ sin x, and from 0 to 2 the value is e² sin 2 = 6.7188.

Takeaway: choose u by LIATE, check that v du is easier than u dv, and remember that some integrals need two rounds or return to themselves.

Class 12 — Tangents and normals, implicit and parametric differentiation

Concept: the tangent at P touches the curve with slope m = dy/dx; the normal is the line through P perpendicular to it, with slope −1 ÷ m. The slope comes from y = f(x) directly, from F(x, y) = 0 by implicit differentiation, or from x(t), y(t) as (dy/dt) ÷ (dx/dt).

Open Tangents & normals, y = x³ − 3x, at x₀ = 2. Then y₀ = 2 and m = 3x² − 3 = 9, so the tangent is y − 2 = 9(x − 2), i.e. 9x − y = 16, and the normal is y − 2 = −(x − 2) ÷ 9, i.e. x + 9y = 20. The tangent meets the x-axis at 16/9 = 1.778.
Lengths. Subtangent = |y₀ ÷ m| = 2/9 = 0.2222 and subnormal = |y₀ m| = 18 (switch on the x-axis marker to see them). Length of tangent = |y₀|√(1 + m²) ÷ |m| = 2.0123 and length of normal = |y₀|√(1 + m²) = 18.11.
Where is the tangent parallel to y = x? Set 3x² − 3 = 1: x = ±2/√3 = ±1.155. The two points are marked on the graph when the slope slider is 1. Horizontal tangents have m = 0 (x = ±1 here).
Implicit. Choose the circle x² + y² = 25 at (3, 4). Differentiating gives 2x + 2y dy/dx = 0, so dy/dx = −x ÷ y = −3/4. The tangent is 3x + 4y = 25 and the normal 4x − 3y = 0 passes through the centre: the normal to a circle is always along a radius. The second derivative −25 ÷ y³ = −0.3906 says the curve is concave down there.
Choose the folium x³ + y³ = 3xy at (1.5, 1.5): dy/dx = −(3x² − 3y) ÷ (3y² − 3x) = −1, so the tangent is x + y = 3. At a point where F_y = 0 the tangent is vertical.
Parametric. For the parabola x = t², y = 2t at t = 2 the point is (4, 4) and dy/dx = 2 ÷ 2t = 1/2, so the tangent is x − 2y + 4 = 0. For the circle (3 cos t, 3 sin t), dy/dx = −cot t = −1 at t = π/4. For the cycloid at t = 2, dy/dx = sin t ÷ (1 − cos t) = 0.6421 and d²y/dx² = (y″x′ − y′x″) ÷ (x′)³ = −1 ÷ (1 − cos t)² = −0.4987.

Takeaway: slope = dy/dx; tangent y − y₀ = m(x − x₀); normal slope −1 ÷ m; implicit dy/dx = −F_x ÷ F_y; parametric dy/dx = y′(t) ÷ x′(t).

The maths behind it

What each mode covers

Limits — type any f(x) and squeeze x towards a from both sides: one-sided limits, infinite limits, oscillation, and whether f is continuous or differentiable there. Class 11 limits and class 12 continuity.

Derivative — a secant you can shrink into a tangent, the difference quotient table, the symbolic derivative, tangent and normal lines and the graph of f′. Class 11 derivatives and class 12 tangents and normals.

Maxima & minima — the curve coloured where it rises and falls, stationary points classified by the first and second derivative tests, points of inflection, concavity and absolute extrema on an interval. Class 12 applications of derivatives.

Optimisation — five classic problems (fenced rectangle, pen against a wall, open box, can of least metal, rectangle in a semicircle) with the four-step working. Class 12 applications of derivatives.

Integrals — Riemann sums with left, right, midpoint and trapezoid rules, the exact integral, the error, signed against total area, and the area function that shows the fundamental theorem. Class 12 integrals.

Differential equations — slope fields, a draggable starting point, Runge–Kutta and Euler solutions and the exact solution for the standard first-order equations. Class 12 differential equations.

Part of Shunya. See the syllabus map for the chapters these modes map to. The Function & Graph Explorer is a good place to look at a function before you differentiate it. Numbers here are computed live from your function, so a few results are numerical estimates and are shown to four decimals.

Formulas used in this lab

Limits Class 11

Definition: $\lim_{x \to a} f(x) = L$
The limit exists when: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$
Algebra of limits: $\lim (f \pm g) = \lim f \pm \lim g,\qquad \lim (fg) = \lim f \cdot \lim g$
Standard limits: $\lim_{x \to 0}\dfrac{\sin x}{x} = 1,\qquad \lim_{x \to 0}\dfrac{e^x - 1}{x} = 1$
More standard limits: $\lim_{x \to 0}\dfrac{\ln(1 + x)}{x} = 1,\qquad \lim_{x \to a}\dfrac{x^n - a^n}{x - a} = na^{n-1}$
The number e: $\lim_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^n = e$

Derivative Class 11

First principles: $f'(x) = \lim_{h \to 0}\dfrac{f(x + h) - f(x)}{h}$
Sum and product: $(u \pm v)' = u' \pm v',\qquad (uv)' = u'v + uv'$
Quotient: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$
Chain rule: $\dfrac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)$
Power and exponential: $(x^n)' = nx^{n-1},\qquad (e^x)' = e^x,\qquad (\ln x)' = \dfrac{1}{x}$
Trigonometric: $(\sin x)' = \cos x,\quad (\cos x)' = -\sin x,\quad (\tan x)' = \sec^2 x$

Continuity and differentiability Class 12

Continuous at a: $\lim_{x \to a} f(x) = f(a)$
Differentiable at a: $\lim_{h \to 0^-}\dfrac{f(a + h) - f(a)}{h} = \lim_{h \to 0^+}\dfrac{f(a + h) - f(a)}{h}$
Differentiable implies continuous: $f'(a) \text{ exists} \Rightarrow f \text{ is continuous at } a$
The reverse is false (|x| is continuous at 0 but not differentiable): $f(x) = |x|$

Tangent and normal Class 12

Slope of the tangent: $m = f'(x_0)$
Tangent line: $y - y_0 = f'(x_0)\,(x - x_0)$
Normal line: $y - y_0 = -\dfrac{1}{f'(x_0)}\,(x - x_0)$
Rate of change of y with respect to x: $\dfrac{dy}{dx}$

Increasing, decreasing, maxima and minima Class 12

Increasing and decreasing: $f'(x) > 0 \Rightarrow \text{increasing},\qquad f'(x) < 0 \Rightarrow \text{decreasing}$
Stationary points: $f'(x) = 0$
First derivative test: $f' \text{ changes } + \to - \Rightarrow \max,\qquad - \to + \Rightarrow \min$
Second derivative test: $f'(c) = 0,\ f''(c) < 0 \Rightarrow \max;\quad f''(c) > 0 \Rightarrow \min$
Concavity and inflection: $f'' > 0:\ \text{concave up},\quad f'' < 0:\ \text{concave down}$
Absolute extrema on [a, b]: compare the ends and every stationary point c: $f(a),\ f(b),\ f(c)$

Optimisation Class 12

Fixed perimeter P: $A(x) = x\left(\dfrac{P}{2} - x\right),\qquad x = \dfrac{P}{4}$
Pen against a wall, fencing L: $A(x) = x(L - 2x),\qquad x = \dfrac{L}{4}$
Open box from a square sheet s: $V(x) = x(s - 2x)^2,\qquad x = \dfrac{s}{6}$
Can of volume V, least metal: $S(r) = 2\pi r^2 + \dfrac{2V}{r},\qquad r = \sqrt[3]{\dfrac{V}{2\pi}},\ h = 2r$
Rectangle in a semicircle of radius R: $A(x) = 2x\sqrt{R^2 - x^2},\qquad x = \dfrac{R}{\sqrt{2}}$
The four steps: $\text{write } f(x);\ f'(x) = 0;\ \text{check } f''(x);\ \text{state the answer}$

Integrals Class 12

Riemann sum: $\sum_{i=1}^{n} f(x_i)\,\Delta x,\qquad \Delta x = \dfrac{b - a}{n}$
Definite integral: $\int_a^b f(x)\,dx = \lim_{n \to \infty}\sum_{i=1}^{n} f(x_i)\,\Delta x$
Fundamental theorem: $\int_a^b f(x)\,dx = F(b) - F(a),\qquad F'(x) = f(x)$
Standard integrals: $\int x^n dx = \dfrac{x^{n+1}}{n + 1},\quad \int e^x dx = e^x,\quad \int \dfrac{dx}{x} = \ln x$
Trigonometric: $\int \sin x\,dx = -\cos x,\qquad \int \cos x\,dx = \sin x$
Properties: $\int_a^b f = -\int_b^a f,\qquad \int_a^b f = \int_a^c f + \int_c^b f$
Trapezoid rule: $\dfrac{\Delta x}{2}\,\bigl[f(x_0) + 2f(x_1) + \dots + f(x_n)\bigr]$

Differential equations Class 12

Order and degree: $\text{order} = \text{highest derivative},\qquad \text{degree} = \text{its power}$
Variable separable: $\dfrac{dy}{dx} = f(x)g(y) \Rightarrow \int \dfrac{dy}{g(y)} = \int f(x)\,dx$
Growth and decay: $\dfrac{dy}{dx} = ky \Rightarrow y = Ce^{kx}$
Linear first order: $\dfrac{dy}{dx} + Py = Q,\qquad \text{integrating factor } e^{\int P\,dx}$
Euler's method: $y_{n+1} = y_n + h\,f(x_n, y_n)$
Particular solution: $\text{use } y(x_0) = y_0 \text{ to find the constant } C$

Integration by substitution and by parts Class 12

Substitution: $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du,\quad u = g(x)$
Definite integral, new limits: $\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$
The form f′ ÷ f: $\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$
Powers of a function: $\int [f(x)]^n f'(x)\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C\ (n \ne -1)$
Integration by parts: $\int u\,dv = uv - \int v\,du$
LIATE (choice of u): $\text{Logarithmic, Inverse trig, Algebraic, Trig, Exponential}$
The e^x shortcut: $\int e^x\,[f(x) + f'(x)]\,dx = e^x f(x) + C$
Cyclic case: $\int e^{ax}\sin bx\,dx = \dfrac{e^{ax}(a\sin bx - b\cos bx)}{a^2 + b^2} + C$

Tangents, normals and implicit or parametric differentiation Class 12

Tangent and normal at (x₀, y₀): $y - y_0 = m(x - x_0),\qquad y - y_0 = -\dfrac{1}{m}(x - x_0),\quad m = \dfrac{dy}{dx}\Big|_{(x_0,y_0)}$
Horizontal and vertical tangent: $\dfrac{dy}{dx} = 0 \ (\text{horizontal}),\qquad \dfrac{dy}{dx} \to \infty \ (\text{vertical})$
Subtangent and subnormal: $\left|\dfrac{y_0}{m}\right|,\qquad |y_0 m|$
Length of tangent and normal: $\left|\dfrac{y_0\sqrt{1+m^2}}{m}\right|,\qquad \left|y_0\sqrt{1+m^2}\right|$
Implicit differentiation: $F(x,y) = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{F_x}{F_y}$
Parametric differentiation: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt},\qquad \dfrac{d^2y}{dx^2} = \dfrac{y''x' - y'x''}{(x')^3}$
Angle between two curves: $\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|$