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# Mathematics I · Week 8 — Derivatives and critical points Deep study for Quiz 2 week 8. Derivative measures instantaneous rate of change; critical points locate where slope is zero or undefined.

Mathematics I · Week 8 — Derivatives and critical points

Deep study for Quiz 2 week 8. Derivative measures instantaneous rate of change; critical points locate where slope is zero or undefined.

Week map

Derivative as slope → differentiation rules → tangent line → critical points → increasing/decreasing → local extrema sketch.

Derivative notation

$f'(x)$ or $\frac{dy}{dx}$ → derivative of $f$ at $x$ → instantaneous rate of change → slope of tangent line at $x$ .
$(x^n)' = nx^{n-1}$ → power rule → bring exponent down, reduce power by one.
$(cf)' = c \cdot f'$ → constant multiple.
$(f + g)' = f' + g'$ → sum rule.

Mini-example: if

f(x) = x^3 - 2x

, then

f'(x) = 3x^2 - 2

. At

x = 1

f'(1) = 1

— tangent slope is 1.

Chain and product (when needed)

Chain: $(f(g(x)))' = f'(g(x)) \cdot g'(x)$ — derivative of outside × derivative of inside.
Product: $(fg)' = f'g + fg'$ .

Mini-example:

(3x^2 + 1)^4

— outer power 4, inner

3x^2+1

. Derivative

= 4(3x^2+1)^3 \cdot 6x

Tangent line

Line through $(a, f(a))$ with slope $m = f'(a)$ : $y - f(a) = f'(a)(x - a)$ .
Tangent approximates curve locally near $x = a$ .

Mini-example:

f(x) = x^2

a = 3

f(3) = 9

f'(x) = 2x

f'(3) = 6

. Tangent:

y - 9 = 6(x - 3)

→

y = 6x - 9

Critical points

Critical point → where $f'(x) = 0$ or $f'(x)$ undefined.
First derivative test: sign of $f'$ changes from $+$ to $-$ at critical point → local max; $-$ to $+$ → local min.
Increasing: $f' > 0$ ; decreasing: $f' < 0$ .

Mini-example:

f(x) = x^2 - 4x + 3

f'(x) = 2x - 4 = 0

→

x = 2

f'

negative before 2, positive after → local min at

x = 2

Pattern families

Easy — Power and sum rule

Differentiate term by term. Constants vanish. Rewrite roots and reciprocals as powers first:

\sqrt{x} = x^{1/2}

1/x = x^{-1}

Medium — Tangent line and evaluate derivative

Find

f'(a)

, write tangent equation. Interpret

f'(a)

as rate (velocity, marginal cost) in word problems.

Hard — Critical points and extrema

Solve

f'(x) = 0

. Build sign chart. Classify max/min. Sketch behavior without full graph.

Worked mini-examples

Example 1 — Power rule.

\frac{d}{dx}(5x^4 - 3x + 7) = 20x^3 - 3

Example 2 — Rewrite then differentiate.

f(x) = \frac{1}{x^2} = x^{-2}

→

f'(x) = -2x^{-3} = -2/x^3

Example 3 — Tangent.

f(x) = x^3

a = 1

f(1) = 1

f'(1) = 3

. Tangent:

y = 3x - 2

Example 4 — Critical point.

f(x) = -x^2 + 6x

f'(x) = -2x + 6 = 0

→

x = 3

. Max (parabola opens down).

Example 5 — Chain rule.

f(x) = (2x + 1)^3

f'(x) = 3(2x+1)^2 \cdot 2 = 6(2x+1)^2

Traps

Forgetting to multiply by inner derivative in chain rule.
Critical point assumed to be max/min without sign test.
Tangent uses wrong point — must pass through $(a, f(a))$ .
$(fg)' written as$ f'g'$ — product rule differs.
Derivative of constant not zero in long expressions.

Diagnostic (try yourself)

Find $f'(x)$ for $f(x) = 4x^3 - 5x^2 + x - 9$ .
Differentiate $g(x) = \sqrt{x} + 1/x$ (rewrite powers first).
Find the equation of the tangent line to $f(x) = x^2$ at $x = -2$ .
For $f(x) = x^3 - 3x$ , find all critical points.
If $f'(x) > 0$ on an interval, is $f$ increasing or decreasing there?