FAC1004 Kahoot Quiz Simulator
Based on your actual quiz archetypes. Every problem below is a direct variant of one of the six types from your notebook.
Target pace: 45–90 seconds per problem (Kahoot speed).
Total: 48 problems.
Do in one sitting. Grade immediately with the answer key.
One-Page Formula Card (Memorize This)
Inverse Hyperbolic Derivatives
| Function | Derivative |
|---|---|
| $\sinh^{-1} u$ | $\frac{u'}{\sqrt{u^2+1}}$ |
| $\cosh^{-1} u$ | $\frac{u'}{\sqrt{u^2-1}}$ $(u>1)$ |
| $\tanh^{-1} u$ | $\frac{u'}{1-u^2}$ $(|u|<1)$ |
| $\coth^{-1} u$ | $\frac{u'}{1-u^2}$ $(|u|>1)$ |
| $\text{sech}^{-1} u$ | $-\frac{u'}{u\sqrt{1-u^2}}$ $(0<u<1)$ |
| $\text{csch}^{-1} u$ | $-\frac{u'}{|u|\sqrt{1+u^2}}$ |
Standard Integrals
$$\int \frac{dx}{\sqrt{a^2+x^2}} = \sinh^{-1}\frac{x}{a} + C$$ $$\int \frac{dx}{\sqrt{x^2-a^2}} = \cosh^{-1}\frac{x}{a} + C$$ $$\int \frac{dx}{a^2-x^2} = \frac{1}{a}\tanh^{-1}\frac{x}{a} + C$$
Chain Rule Reminder
If $y = f(g(x))$, then $y' = f'(g(x)) \cdot g'(x)$.
If $y = uv$, then $y' = u'v + uv'$.
Archetype A — Derivative of Inverse Hyperbolic (Linear / Polynomial Inner)
Pattern: $\frac{d}{dx}[\text{inv-hyp}(f(x))]$
Move: Apply formula, then multiply by $f'(x)$.
- $y = \sinh^{-1}(3x+1)$
- $y = \cosh^{-1}(2x-5)$
- $y = \tanh^{-1}(4x)$
- $y = \sinh^{-1}(x^2)$
- $y = \text{sech}^{-1}(5x)$
- $y = \coth^{-1}(x^3+2)$
- $y = \text{csch}^{-1}(\sqrt{x})$
- $y = \cosh^{-1}\left(\frac{x}{3}\right)$
Archetype B — Derivative of Inverse Hyperbolic (Exponential / Trig Inner)
Pattern: $\frac{d}{dx}[\text{inv-hyp}(e^{kx})]$ or $\text{inv-hyp}(\sin x, \cos x, \ln x)$
Move: Let $u = \text{inner function}$, apply formula, include $u'$.
- $y = \text{sech}^{-1}(e^{3x})$
- $y = \sinh^{-1}(e^{2x})$
- $y = \cosh^{-1}(e^{x/2})$
- $y = \tanh^{-1}(\sin x)$
- $y = \sinh^{-1}(\ln x)$
- $y = \text{sech}^{-1}(e^{-x})$
- $y = \coth^{-1}(\cos x)$
- $y = \cosh^{-1}(e^{4x}+1)$
Archetype C — Product Rule: $e^{ax} \times$ Inverse Hyperbolic
Pattern: $\frac{d}{dx}\left[e^{ax} \cdot \sinh^{-1}\left(\frac{b}{x^n}\right)\right]$
Move: $u = e^{ax}$, $v = \text{inv-hyp}(\dots)$. Use $(uv)' = u'v + uv'$.
- $y = e^{3x} \sinh^{-1}\left(\frac{2}{x}\right)$
- $y = e^{x} \cosh^{-1}(x^2+1)$
- $y = e^{-2x} \tanh^{-1}(\sqrt{x})$
- $y = e^{5x} \sinh^{-1}\left(\frac{4}{x^3}\right)$
- $y = x^2 \cosh^{-1}(3x)$
- $y = e^{x/2} \text{sech}^{-1}(x^2)$
- $y = x \sinh^{-1}(e^{x})$
- $y = e^{4x} \coth^{-1}\left(\frac{1}{x}\right)$
Archetype D — Hyperbolic U-Substitution Integrals
Pattern: $\int \cosh^n(x)\sinh(x),dx$ or $\int \sinh^n(x)\cosh(x),dx$
Move: Let $u = \cosh x$ (if $\sinh x,dx$ present) or $u = \sinh x$ (if $\cosh x,dx$ present).
- $\int \cosh^5(x) \sinh(x),dx$
- $\int \sinh^3(x) \cosh(x),dx$
- $\int \cosh^4(x) \sinh(x),dx$
- $\int \sinh^5(x) \cosh(x),dx$
- $\int \cosh^2(x) \sinh(x),dx$
- $\int \sinh^2(x) \cosh(x),dx$
- $\int \text{sech}^2(x) \tanh(x),dx$
- $\int \text{csch}^2(x) \coth(x),dx$
Archetype E — Linear Substitution Integrals
Pattern: $\int f(ax+b),dx$
Move: Let $u = ax+b$, $du = a,dx$, so $dx = \frac{du}{a}$.
- $\int \frac{dx}{3-6x}$
- $\int e^{4-8x},dx$
- $\int \sin(2x+5),dx$
- $\int \frac{dx}{5x+2}$
- $\int \cosh(3x-1),dx$
- $\int \sinh\left(\frac{x}{2}+3\right),dx$
- $\int \text{sech}^2(4x),dx$
- $\int \frac{dx}{2-10x}$
Archetype F — Standard Form → Inverse Hyperbolic
Pattern: $\int \frac{dx}{\sqrt{a^2+b^2x^2}}$, $\int \frac{dx}{\sqrt{x^2-a^2}}$, $\int \frac{dx}{a^2-x^2}$
Move: Factor out constants to match standard form exactly.
- $\int \frac{dx}{\sqrt{4+9x^2}}$
- $\int \frac{dx}{\sqrt{1+16x^2}}$
- $\int \frac{dx}{\sqrt{x^2-25}}$
- $\int \frac{dx}{\sqrt{9x^2-4}}$
- $\int \frac{dx}{9-4x^2}$
- $\int \frac{dx}{25-x^2}$
- $\int \frac{dx}{\sqrt{4x^2+1}}$
- $\int \frac{dx}{\sqrt{(x+1)^2+9}}$ (complete the square first)
Answer Key
| # | Answer | # | Answer |
|---|---|---|---|
| 1 | $\frac{3}{\sqrt{(3x+1)^2+1}}$ | 25 | $\frac{\cosh^6 x}{6} + C$ |
| 2 | $\frac{2}{\sqrt{(2x-5)^2-1}}$ | 26 | $\frac{\sinh^4 x}{4} + C$ |
| 3 | $\frac{4}{1-16x^2}$ | 27 | $\frac{\cosh^5 x}{5} + C$ |
| 4 | $\frac{2x}{\sqrt{x^4+1}}$ | 28 | $\frac{\sinh^6 x}{6} + C$ |
| 5 | $-\frac{1}{x\sqrt{1-25x^2}}$ | 29 | $\frac{\cosh^3 x}{3} + C$ |
| 6 | $\frac{3x^2}{1-(x^3+2)^2}$ | 30 | $\frac{\sinh^3 x}{3} + C$ |
| 7 | $-\frac{1}{2x\sqrt{x+1}}$ (or $-\frac{1}{2x^{3/2}\sqrt{1+1/x}}$) | 31 | $-\frac{\text{sech}^2 x}{2} + C$ or $\frac{\tanh^2 x}{2} + C$ |
| 8 | $\frac{1}{\sqrt{x^2-9}}$ | 32 | $-\frac{\coth^2 x}{2} + C$ |
| 9 | $-\frac{3}{\sqrt{1-e^{6x}}}$ | 33 | $-\frac{1}{6}\ln|3-6x| + C$ |
| 10 | $\frac{2e^{2x}}{\sqrt{e^{4x}+1}}$ | 34 | $-\frac{1}{8}e^{4-8x} + C$ |
| 11 | $\frac{e^{x/2}}{2\sqrt{e^x-1}}$ | 35 | $-\frac{1}{2}\cos(2x+5) + C$ |
| 12 | $\frac{\cos x}{1-\sin^2 x} = \sec x$ | 36 | $\frac{1}{5}\ln|5x+2| + C$ |
| 13 | $\frac{1}{x\sqrt{(\ln x)^2+1}}$ | 37 | $\frac{1}{3}\sinh(3x-1) + C$ |
| 14 | $\frac{1}{\sqrt{1-e^{-2x}}}$ | 38 | $2\cosh\left(\frac{x}{2}+3\right) + C$ |
| 15 | $\frac{\sin x}{1-\cos^2 x} = \csc x$ | 39 | $\frac{1}{4}\tanh(4x) + C$ |
| 16 | $\frac{4e^{4x}}{\sqrt{(e^{4x}+1)^2-1}}$ | 40 | $-\frac{1}{10}\ln|2-10x| + C$ |
| 17 | $3e^{3x}\sinh^{-1}\left(\frac{2}{x}\right) + e^{3x}\cdot\frac{-2/x^2}{\sqrt{4/x^2+1}}$ | 41 | $\frac{1}{3}\sinh^{-1}\left(\frac{3x}{2}\right) + C$ |
| 18 | $e^x\cosh^{-1}(x^2+1) + e^x\cdot\frac{2x}{\sqrt{(x^2+1)^2-1}}$ | 42 | $\frac{1}{4}\sinh^{-1}(4x) + C$ |
| 19 | $-2e^{-2x}\tanh^{-1}(\sqrt{x}) + e^{-2x}\cdot\frac{1}{2\sqrt{x}(1-x)}$ | 43 | $\cosh^{-1}\left(\frac{x}{5}\right) + C$ |
| 20 | $5e^{5x}\sinh^{-1}\left(\frac{4}{x^3}\right) + e^{5x}\cdot\frac{-12/x^4}{\sqrt{16/x^6+1}}$ | 44 | $\frac{1}{3}\cosh^{-1}\left(\frac{3x}{2}\right) + C$ |
| 21 | $2x\cosh^{-1}(3x) + x^2\cdot\frac{3}{\sqrt{9x^2-1}}$ | 45 | $\frac{1}{6}\tanh^{-1}\left(\frac{2x}{3}\right) + C$ |
| 22 | $\frac{1}{2}e^{x/2}\text{sech}^{-1}(x^2) + e^{x/2}\cdot\frac{-2}{x\sqrt{1-x^4}}$ | 46 | $\frac{1}{5}\tanh^{-1}\left(\frac{x}{5}\right) + C$ |
| 23 | $\sinh^{-1}(e^x) + x\cdot\frac{e^x}{\sqrt{e^{2x}+1}}$ | 47 | $\frac{1}{2}\sinh^{-1}(2x) + C$ |
| 24 | $4e^{4x}\coth^{-1}\left(\frac{1}{x}\right) + e^{4x}\cdot\frac{1}{1-1/x^2}\cdot\left(-\frac{1}{x^2}\right)$ | 48 | $\sinh^{-1}\left(\frac{x+1}{3}\right) + C$ |
Simplified forms for common answers
- #12: $\frac{\cos x}{\cos^2 x} = \sec x$
- #15: $\frac{\sin x}{\sin^2 x} = \csc x$
- #17 simplified: $3e^{3x}\sinh^{-1}\frac{2}{x} - \frac{2e^{3x}}{x\sqrt{x^2+4}}$
- #20 simplified: $5e^{5x}\sinh^{-1}\frac{4}{x^3} - \frac{12e^{5x}}{x\sqrt{x^6+16}}$
- #24 simplified: $4e^{4x}\coth^{-1}\frac{1}{x} + \frac{e^{4x}}{x^2-1}$
Scoring
| Archetype | Problems | Your Score |
|---|---|---|
| A — Basic Inv-Hyp Derivatives | 1–8 | ___/8 |
| B — Inv-Hyp with Exp/Trig Inner | 9–16 | ___/8 |
| C — Product Rule (Exp × Inv-Hyp) | 17–24 | ___/8 |
| D — Hyperbolic U-Sub Integrals | 25–32 | ___/8 |
| E — Linear Substitution | 33–40 | ___/8 |
| F — Standard Form Integrals | 41–48 | ___/8 |
| TOTAL | 48 | ___/48 |
Kahoot Readiness Scale
- ≥ 40/48 (83%): You will dominate the Kahoot.
- ≥ 34/48 (71%): You'll keep up, but review missed archetypes.
- < 34/48: Drill the weakest archetype again before the quiz.
5-Minute Blitz Protocol
If you only have 5 minutes before the quiz, do this:
- Recite the 6 inverse hyperbolic derivative formulas (30 seconds).
- Do problems 1, 9, 17 (2 minutes) — one from each derivative archetype.
- Do problems 25, 33, 41 (2 minutes) — one from each integral archetype.
- Check answers. If all correct, you're warmed up. If any wrong, re-memorize that formula.
Related Resources
- FAC1004 Tutorial 7 — Derivatives of Inverse Trig & Hyperbolic Functions
- FAC1004 Tutorial 8 — Hyperbolic Functions
- FAC1004 Tutorial 9 — Inverse Hyperbolic Functions
- FAC1004 Rapid-Fire Drill Pack — Inverse Trig, Hyperbolic & Inverse Hyperbolic
- FAC1004 - Advanced Mathematics II (Computing)
#mathematics #kahoot #quiz-prep #inverse-hyperbolic #hyperbolic-integrals #fac1004 #timed-drill