📐 Calculus Self-Study Notebook

Chain Rule · Integration by Substitution · Plugin Method

✏️ Fill in, solve, and check — your pace, your notes!

Part 1 — Concept Review

🔗 A. Chain Rule (Differentiation)

📖 THE BIG IDEA

When you differentiate a function inside another function, you must multiply by the derivative of the inside.

\(\dfrac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)

🧠

MEMORY POINT — say it aloud!

"OUTSIDE stays, INSIDE stays, multiply inside's derivative"
→ derivative of OUTSIDE × derivative of INSIDE

✏️ WORKED EXAMPLE A1

Find \(\dfrac{d}{dx}[\sin(3x^2)]\)

→ Outside function: \(\sin(\square)\), Inside function: \(3x^2\)

→ Derivative of outside: \(\cos(\square)\) → keep inside: \(\cos(3x^2)\)

→ Derivative of inside: \(\dfrac{d}{dx}[3x^2] = 6x\)

→ Multiply: \(\cos(3x^2) \cdot 6x\) = 6x\cos(3x^2) ✓

⚠ COMMON MISTAKE: Forgetting to multiply by the derivative of the inside! ⚠

🔌 B. Integration by Substitution (Plugin / u-sub)

📖 THE BIG IDEA

Substitution is the reverse of the Chain Rule. We "plugin" \(u\) to simplify a messy integral.

\(\displaystyle\int f(g(x))\cdot g'(x)\,dx = \int f(u)\,du\) where \(u = g(x)\)

🔌

MEMORY POINT — 4-step plugin method

1. SPOT the inside function → set \(u\)
2. FIND \(du = u'\,dx\) → solve for \(dx\)
3. SWAP everything to \(u\)
4. INTEGRATE, then swap \(u\) back to \(x\)

✏️ WORKED EXAMPLE B1

Find \(\displaystyle\int 2x\cos(x^2)\,dx\)

→ SPOT: Inside = \(x^2\), so let \(u = x^2\)

→ FIND: \(du = 2x\,dx\) → \(dx = \dfrac{du}{2x}\)

→ SWAP: \(\displaystyle\int \cancel{2x}\cos(u)\dfrac{du}{\cancel{2x}} = \int\cos(u)\,du\)

→ INTEGRATE: \(\sin(u) + C\)

→ BACK: \(\sin(x^2) + C\) ✓

⚡ KEY WORDS to look for

SPOT the pattern: Is there an "inside function" and its derivative sitting next to each other?
ADJUST constants: If you're off by a constant factor, you can fix it with multiplication.
DEFINITE integral: Change the limits when you change variables! \(x=a \Rightarrow u=g(a)\)

Part 2 — Chain Rule Practice (Q1–Q10)

Choose the best answer. ✓ = correct, ✗ = see explanation!

0 / 20 answered

Q 01 — Chain Rule Easy ★☆☆

Find \(\dfrac{d}{dx}[(3x+1)^5]\)

Outside = power of 5, Inside = \(3x+1\)

Q 02 — Chain Rule Easy ★☆☆

Find \(\dfrac{d}{dx}[e^{4x}]\)

Remember: \(\dfrac{d}{dx}[e^u] = e^u \cdot u'\)

Q 03 — Chain Rule Easy ★☆☆

Find \(\dfrac{d}{dx}[\sin(5x)]\)

Derivative of \(\sin\) is \(\cos\), then multiply inside's derivative

Q 04 — Chain Rule Easy ★☆☆

Find \(\dfrac{d}{dx}[\ln(x^2+1)]\)

\(\dfrac{d}{dx}[\ln u] = \dfrac{u'}{u}\)

Q 05 — Chain Rule Tricky ★★☆

Find \(\dfrac{d}{dx}[\cos^3(x)]\)
(i.e., \([\cos(x)]^3\))

Two chain rules here! Outside = cube, Inside = \(\cos(x)\)

Q 06 — Chain Rule Tricky ★★☆

Find \(\dfrac{d}{dx}\!\left[\sqrt{x^3+2}\right]\)

Rewrite: \((x^3+2)^{1/2}\), then chain rule with power rule

Q 07 — Chain Rule Tricky ★★☆

Find \(\dfrac{d}{dx}\!\left[e^{x^2+3x}\right]\)

Inside function is \(x^2+3x\), find its derivative first

Q 08 — Chain Rule Hard ★★★

Find \(\dfrac{d}{dx}\!\left[\tan(e^x)\right]\)

Outer = \(\tan\), Inner = \(e^x\). Recall \(\dfrac{d}{dx}[\tan u] = \sec^2(u)\cdot u'\)

Q 09 — Chain Rule Hard ★★★

If \(y = \ln(\sin x)\), find \(\dfrac{dy}{dx}\)

Outer = \(\ln\), Inner = \(\sin x\)

Q 10 — Chain Rule Hard ★★★

Find \(\dfrac{d}{dx}\!\left[\sin^2(3x)\right]\)

Three layers! Power rule → sin → 3x

Part 3 — Integration by Substitution (Q11–Q20)

🔑

QUICK REMINDER — u-substitution steps

SPOT → SET → SWAP → INTEGRATE → BACK

Q 11 — u-sub Easy ★☆☆

Find \(\displaystyle\int (2x+1)^3 \cdot 2\,dx\)

Let \(u = 2x+1\), then \(du = 2\,dx\)

Q 12 — u-sub Easy ★☆☆

Find \(\displaystyle\int \cos(3x)\,dx\)

Let \(u = 3x\), so \(du = 3\,dx\) → you need to adjust by a factor of \(\frac{1}{3}\)

Q 13 — u-sub Easy ★☆☆

Find \(\displaystyle\int e^{5x}\,dx\)

Let \(u = 5x\)

Q 14 — u-sub Easy ★☆☆

Find \(\displaystyle\int \dfrac{1}{x+4}\,dx\)

Let \(u = x+4\), recall \(\int \frac{1}{u}\,du = \ln|u|+C\)

Q 15 — u-sub Tricky ★★☆

Find \(\displaystyle\int 3x^2 e^{x^3}\,dx\)

The key: \(3x^2\) is the derivative of \(x^3\) — perfect setup!

Q 16 — u-sub Tricky ★★☆

Find \(\displaystyle\int \sin(x)\cos(x)\,dx\)

Let \(u = \sin(x)\), then \(du = \cos(x)\,dx\) — check if everything cancels nicely!

Q 17 — u-sub (Definite) Tricky ★★☆

Evaluate \(\displaystyle\int_0^1 2x(x^2+1)^3\,dx\)

Let \(u = x^2+1\). Change limits! \(x=0 \Rightarrow u=1\), \(x=1 \Rightarrow u=2\)

Q 18 — u-sub Tricky ★★☆

Find \(\displaystyle\int \dfrac{2x}{x^2+5}\,dx\)

Notice: numerator \(2x\) is the derivative of denominator \(x^2+5\)!

Q 19 — u-sub Hard ★★★

Find \(\displaystyle\int x\sqrt{x^2-4}\,dx\)

Let \(u = x^2-4\), so \(du = 2x\,dx\) → factor out \(\frac{1}{2}\)

Q 20 — u-sub (Chain Reversed) Hard ★★★

Find \(\displaystyle\int \cos^4(x)\sin(x)\,dx\)

Let \(u = \cos(x)\), then \(du = -\sin(x)\,dx\) — watch the negative sign!