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Mathematical Architects · Calculus · Foundations

Get Ready: Module 2 — Derivatives: Definition & Techniques

Everyone starts somewhere. Before we measure slope at a point, we'll firm up the four ideas the derivative leans on — clearly, one step at a time, with nothing assumed. Go at your own pace. There's no rush and no wrong place to begin.

4 Foundation Skills Worked Examples Self-Paced

The derivative is built from a limit — "what happens as the gap \(h\) shrinks to \(0\)" — applied to composed functions you'll need to recognize, including exponentials, logs, and trig. If limits, \(f(g(x))\), exponent rules, or the unit-circle values feel fuzzy right now, that's completely okay — we'll rebuild them together below, then you'll be ready for the lab.

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Skills to build first

Four short skills. Read the idea, follow the worked example, then try one yourself — the answer is one click away whenever you want it.

1 Skill

Limits and the idea of \(h \to 0\)

What it is: a limit asks what value an expression approaches as the input gets arbitrarily close to a target — even if you can't plug the target in directly. The derivative's engine, \(\frac{f(x+h)-f(x)}{h}\), is \(\frac00\) if you set \(h=0\) outright, so we simplify first, then let \(h \to 0\). Why you need it: the entire definition of the derivative is a limit. This is Module 1 material — the direct prerequisite for Module 2.

Worked example — evaluate \(\displaystyle\lim_{h\to 0}\frac{(2+h)^2-4}{h}\)
  1. Don't plug in yet. At \(h=0\) this is \(\frac{0}{0}\) — indeterminate, not the answer.
  2. Expand the top. \((2+h)^2-4 = 4 + 4h + h^2 - 4 = 4h + h^2\).
  3. Factor and cancel the \(h\). \(\frac{4h+h^2}{h} = \frac{h(4+h)}{h} = 4 + h\).
  4. Now let \(h\to 0\). \(4 + h \to 4\). So the limit is \(4\).

That number, \(4\), is exactly the slope of \(y=x^2\) at \(x=2\) — a derivative, computed by a limit.

Try it 1 — Evaluate \(\displaystyle\lim_{h\to 0}\frac{(3+h)^2-9}{h}\).
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Top: \((3+h)^2-9 = 6h + h^2\). Cancel \(h\): \(6 + h\). Let \(h\to 0\): the limit is \(\mathbf{6}\).

Try it 2 — Evaluate \(\displaystyle\lim_{h\to 0}\frac{5h + h^2}{h}\).
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Factor: \(\frac{h(5+h)}{h} = 5 + h\). As \(h\to 0\), the value is \(\mathbf{5}\).

Practice more (free) Khan Academy → IXL →
2 Skill

Function composition \(f(g(x))\) & the function families

What it is: a composition \(f(g(x))\) feeds one function's output into another — an "outer" function wrapped around an "inner" one. You also need to recognize the core families: powers \(x^n\), \(\sqrt{x}\), \(\sin x\) and \(\cos x\), \(e^x\), and \(\ln x\). Why you need it: the chain rule — the rule everyone forgets — is entirely about spotting the outer and inner functions in a composition. If you can decompose \(\sin(3x)\) into "sine of (three x)," the chain rule is easy. (Composition & families are Pre-Calculus prerequisites.)

Worked example — identify outer and inner in \(\sqrt{x^2+1}\)
  1. Ask "what's done last?" You square-and-add first, then take the square root. The square root is the outer function.
  2. Name the inner. The stuff under the root, \(g(x)=x^2+1\), is the inner function.
  3. Write it as a composition. With \(f(u)=\sqrt{u}\), we have \(\sqrt{x^2+1} = f(g(x))\).
  4. Check by evaluating. At \(x=2\): inner \(g(2)=5\), outer \(\sqrt{5}\). Correct.

This "outer of inner" reading is exactly what the chain rule needs: differentiate the outer, then multiply by the inner's derivative.

Try it 1 — If \(f(x)=x^2\) and \(g(x)=x+3\), find \(f(g(x))\) and \(g(f(x))\).
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\(f(g(x)) = (x+3)^2\) (square the inner). \(g(f(x)) = x^2 + 3\) (add 3 to the inner). Order matters!

Try it 2 — Decompose \(e^{2x}\) into an outer and an inner function.
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Outer \(f(u)=e^{u}\); inner \(g(x)=2x\). So \(e^{2x} = f(g(x))\). (The inner's derivative, \(2\), is the chain-rule factor.)

Practice more (free) Khan Academy → IXL →
3 Skill

Exponent & logarithm rules

What it is: the algebra of powers (\(x^a\cdot x^b = x^{a+b}\), \(\;(x^a)^b = x^{ab}\), \(\;x^{-n}=\frac{1}{x^n}\), \(\;\sqrt[n]{x}=x^{1/n}\)) and of logs (\(\ln(ab)=\ln a + \ln b\), \(\;\ln(x^p)=p\ln x\)). Why you need it: the power rule \(\frac{d}{dx}x^n = nx^{n-1}\) only works once you've rewritten roots and reciprocals as powers — \(\sqrt{x}=x^{1/2}\), \(\frac{1}{x}=x^{-1}\). And derivatives of \(e^x\) and \(\ln x\) ride on these rules.

Worked example — rewrite as powers of \(x\)
  1. \(\sqrt{x} = x^{1/2}\) — a root is a fractional power.
  2. \(\dfrac{1}{x^3} = x^{-3}\) — a reciprocal is a negative power.
  3. \(\dfrac{1}{\sqrt{x}} = x^{-1/2}\) — combine both ideas.
  4. \(x^2 \cdot x^5 = x^{7}\) — add exponents when multiplying like bases.

Once everything is \(x^n\), the power rule applies directly: e.g. \(\frac{d}{dx}x^{1/2} = \tfrac12 x^{-1/2}\).

Try it 1 — Rewrite \(\dfrac{1}{x^4}\) and \(\sqrt[3]{x}\) as powers of \(x\).
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\(\dfrac{1}{x^4} = x^{-4}\); \(\;\sqrt[3]{x} = x^{1/3}\).

Try it 2 — Use a log rule to expand \(\ln(x^5)\).
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The power comes down in front: \(\ln(x^5) = 5\ln x\).

Practice more (free) Khan Academy → IXL →
4 Skill

Exact trig values from the unit circle

What it is: the sine and cosine of the common angles — \(0,\ \frac{\pi}{6},\ \frac{\pi}{4}, \ \frac{\pi}{3},\ \frac{\pi}{2}\) — read straight off the unit circle, plus the fact that \(\sin\) and \(\cos\) oscillate between \(-1\) and \(1\). Why you need it: Module 2 establishes \(\frac{d}{dx}\sin x = \cos x\) and \(\frac{d}{dx}\cos x = -\sin x\). To check a tangent slope or evaluate a derivative at a specific angle, you need these values cold (in radians — calculus always uses radians).

Worked example — key values to know
  1. \(\sin 0 = 0\), \(\;\cos 0 = 1\) — start of the circle, on the positive \(x\)-axis.
  2. \(\sin\frac{\pi}{6} = \tfrac12\), \(\;\cos\frac{\pi}{6} = \tfrac{\sqrt3}{2}\) (that's \(30^\circ\)).
  3. \(\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \tfrac{\sqrt2}{2}\) (that's \(45^\circ\) — equal legs).
  4. \(\sin\frac{\pi}{2} = 1\), \(\;\cos\frac{\pi}{2} = 0\) — top of the circle.

So, e.g., the derivative of \(\sin x\) at \(x=0\) is \(\cos 0 = 1\) — the sine curve climbs with slope \(1\) at the origin.

Try it 1 — What are \(\cos\frac{\pi}{3}\) and \(\sin\frac{\pi}{3}\)?
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\(\cos\frac{\pi}{3} = \tfrac12\); \(\;\sin\frac{\pi}{3} = \tfrac{\sqrt3}{2}\) (that's \(60^\circ\)).

Try it 2 — Since \(\frac{d}{dx}\sin x = \cos x\), what is the slope of \(\sin x\) at \(x=\frac{\pi}{2}\)?
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Slope \(= \cos\frac{\pi}{2} = 0\). At the top of the wave the sine curve is momentarily flat — a maximum. (You'll see exactly this in the lab.)

Practice more (free) Khan Academy → IXL →
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Quick Readiness Check

Six short questions spanning all four skills. Try each one first, then reveal the answer to check yourself.

  1. Evaluate \(\displaystyle\lim_{h\to 0}\frac{(4+h)^2-16}{h}\).
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    Top: \(8h+h^2\). Cancel \(h\): \(8+h\). Limit \(= 8\).

  2. If \(f(x)=\sqrt{x}\) and \(g(x)=x+1\), write \(f(g(x))\).
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    \(f(g(x)) = \sqrt{x+1}\) (substitute the inner into the root).

  3. Rewrite \(\dfrac{1}{\sqrt{x}}\) as a power of \(x\).
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    \(\dfrac{1}{\sqrt{x}} = x^{-1/2}\).

  4. Expand \(\ln(3x)\) using a log rule.
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    \(\ln(3x) = \ln 3 + \ln x\) (product rule for logs).

  5. What is \(\cos\frac{\pi}{4}\)?
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    \(\cos\frac{\pi}{4} = \tfrac{\sqrt2}{2}\) (the \(45^\circ\) value).

  6. Decompose \(\cos(5x)\) into an outer and an inner function.
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    Outer \(\cos(u)\); inner \(5x\). The inner's derivative, \(5\), will be the chain-rule factor.

If these feel comfortable, you're ready for the module. If one or two felt shaky, that's your cue — scroll back up to that skill, re-read the worked example, and try a couple from the free practice links. No score to hit, no clock running. Come back when it clicks.

Floor poured. When the skills above feel steady, step into the module — the Visual Lab lets you watch a secant line become a tangent and build the derivative with your own hands.

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