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maths1-week8
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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 dxdy → derivative of f at x → instantaneous rate of change → slope of tangent line at x.
- (xn)′=nxn−1 → power rule → bring exponent down, reduce power by one.
- (cf)′=c⋅f′ → constant multiple.
- (f+g)′=f′+g′ → sum rule.
Mini-example: if f(x)=x3−2x, then f′(x)=3x2−2. At x=1: f′(1)=1 — tangent slope is 1.
Chain and product (when needed)
- Chain: (f(g(x)))′=f′(g(x))⋅g′(x) — derivative of outside × derivative of inside.
- Product: (fg)′=f′g+fg′.
Mini-example: (3x2+1)4 — outer power 4, inner 3x2+1. Derivative =4(3x2+1)3⋅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)=x2, 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)=x2−4x+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: x=x1/2, 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.
dxd(5x4−3x+7)=20x3−3.
Example 2 — Rewrite then differentiate.
f(x)=x21=x−2 → f′(x)=−2x−3=−2/x3.
Example 3 — Tangent.
f(x)=x3, a=1. f(1)=1, f′(1)=3. Tangent: y=3x−2.
Example 4 — Critical point.
f(x)=−x2+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⋅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)′writtenasf'g'$ — product rule differs.
- Derivative of constant not zero in long expressions.
Diagnostic (try yourself)
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Find f′(x) for f(x)=4x3−5x2+x−9.
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Differentiate g(x)=x+1/x (rewrite powers first).
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Find the equation of the tangent line to f(x)=x2 at x=−2.
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For f(x)=x3−3x, find all critical points.
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If f′(x)>0 on an interval, is f increasing or decreasing there?