// CORE 20 PROBLEMS · SELF-STUDY EDITION · ENGLISH
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When a limit gives the indeterminate forms 0/0 or ∞/∞, take the derivative of numerator and denominator separately, then evaluate again.
Find: lim(x→0) (sin 3x) / (5x)
Step 1 — Check form: sin(0)/0 = 0/0 ✓ → apply L'Hôpital
Step 2 — Differentiate top & bottom: 3cos(3x) / 5
Step 3 — Plug in x = 0: 3cos(0)/5 = 3/5
The chain rule handles functions inside functions. Always identify the outer and inner function first, then multiply their derivatives.
[f(g(x))]' = f '(g(x)) · g'(x)
Differentiate: y = sin(e^(3x))
Outer → sin( ), Inner → e^(3x)
dy/dx = cos(e^(3x)) · e^(3x) · 3 = 3e^(3x) cos(e^(3x))
When y is not isolated, differentiate both sides with respect to x. Every time you differentiate a y-term, multiply by dy/dx (chain rule on y).
Find dy/dx: x³ + y³ = 6xy
Step 1 differentiate both sides: 3x² + 3y²·(dy/dx) = 6y + 6x·(dy/dx)
Step 2 collect dy/dx terms: 3y²(dy/dx) − 6x(dy/dx) = 6y − 3x²
Step 3 factor & solve: dy/dx = (6y − 3x²)/(3y² − 6x) = (2y−x²)/(y²−2x)
FTC Part 1 says: d/dx [∫ₐˣ f(t) dt] = f(x). But when the upper limit is a function g(x), chain rule applies!
d/dx [∫ₐ^(g(x)) f(t) dt] = f(g(x)) · g'(x)
Find: d/dx [∫₀^(x²) sin(t) dt]
Step 1 Upper limit = x², so g(x) = x², g'(x) = 2x
Step 2 Apply formula: sin(x²) · 2x = 2x sin(x²)
Formula: ∫u dv = uv − ∫v du
Choose u using LIATE: Logarithmic → Inverse trig → Algebraic → Trig → Exponential. The first matching type is u.
Compute: ∫ x·eˣ dx
LIATE: x is Algebraic (A), eˣ is Exponential (E) → pick u = x, dv = eˣ dx
du = dx, v = eˣ
= x·eˣ − ∫eˣ dx = xeˣ − eˣ + C
Break a rational function with a factorable denominator into simpler fractions before integrating.
Compute: ∫ 5/(x²−x−6) dx
Step 1 Factor denom: (x−3)(x+2)
Step 2 Set up: 5/[(x−3)(x+2)] = A/(x−3) + B/(x+2)
Step 3 Multiply through: 5 = A(x+2)+B(x−3). Plug x=3: A=1; x=−2: B=−1
Step 4 Integrate: ln|x−3| − ln|x+2| + C
Replace ∞ with a limit variable t, evaluate the definite integral, then take the limit as t → ∞. If the limit exists = converges; if not = diverges.
Does ∫₁^∞ (1/x²) dx converge?
= lim(t→∞) ∫₁ᵗ x⁻² dx = lim(t→∞) [−x⁻¹]₁ᵗ = lim(t→∞) (−1/t + 1) = 1 ✓ Converges
A geometric series Σ a·rⁿ (n=0 to ∞):
Sum = a / (1 − r), converges ONLY when |r| < 1
Find the sum: Σ (2/3)ⁿ from n=0 to ∞
a = 1, r = 2/3. |r| = 2/3 < 1 → converges
Sum = 1/(1 − 2/3) = 1/(1/3) = 3
Compute L = lim(n→∞) |a(n+1) / aₙ|
If L < 1 → converges absolutely | L > 1 → diverges | L = 1 → inconclusive
Test: Σ n!/3ⁿ
|a(n+1)/aₙ| = [(n+1)!/3^(n+1)] · [3ⁿ/n!] = (n+1)/3 → ∞
L = ∞ > 1 → Diverges
A Taylor series centered at a: f(x) = Σ f⁽ⁿ⁾(a)/n! · (x−a)ⁿ
Maclaurin = Taylor centered at a = 0. These four are the most tested — memorize them:
Find the Maclaurin series for f(x) = e^(−x²)
Start from eˣ = Σ xⁿ/n!, substitute x → −x²:
e^(−x²) = Σ (−x²)ⁿ/n! = Σ (−1)ⁿ x^(2n)/n!
The error of the nth-degree Taylor polynomial satisfies:
|Error| ≤ M · |x−a|^(n+1) / (n+1)!
where M is the maximum value of |f^(n+1)| on the interval between a and x.
Use a 3rd-degree Maclaurin polynomial for sin(x) to estimate sin(0.1). Find the error bound.
f⁽⁴⁾(x) = sin(x) → |M| ≤ 1 on [0, 0.1]
|Error| ≤ 1 · (0.1)⁴ / 4! = 0.0001/24 ≈ 4.17 × 10⁻⁶
Use the Ratio Test to find where a power series converges. Set L < 1 and solve for x. Then check the endpoints separately!
Find the IOC for Σ (x−2)ⁿ / (n·3ⁿ)
Ratio test: |(x−2)/3| · n/(n+1) → |x−2|/3 < 1 → |x−2| < 3
Radius = 3. Center = 2. Potential IOC: (−1, 5)
Check endpoints: x=−1 → Σ(−1)ⁿ/n → converges (alternating). x=5 → Σ1/n → diverges.
IOC = [−1, 5)
dy/dx = (dy/dt) / (dx/dt)
d²y/dx² = [d/dt(dy/dx)] / (dx/dt)
Note: do not just take (d²y/dt²)/(d²x/dt²) — that is WRONG!
x = t², y = t³. Find dy/dx and d²y/dx².
dx/dt = 2t, dy/dt = 3t²
dy/dx = 3t²/2t = 3t/2
d/dt(3t/2) = 3/2 → d²y/dx² = (3/2)/(2t) = 3/(4t)
L = ∫ₐᵇ √[(dx/dt)² + (dy/dt)²] dt
Find the arc length of x = 3t, y = 4t for t ∈ [0, 1].
dx/dt = 3, dy/dt = 4
L = ∫₀¹ √(9 + 16) dt = ∫₀¹ 5 dt = 5
A = (1/2) ∫ₐᵇ [r(θ)]² dθ
For the area between two polar curves: A = (1/2)∫[r_outer² − r_inner²] dθ
Find the area enclosed by r = 2cos(θ)
This is a circle. It goes from θ = −π/2 to π/2 (one full loop).
A = (1/2) ∫_{−π/2}^{π/2} (2cosθ)² dθ = (1/2)∫4cos²θ dθ = 2∫½(1+cos2θ)dθ
= [θ + sin(2θ)/2]₋π/₂^π/₂ = π
A separable DE has the form dy/dx = f(x)·g(y). Separate variables, integrate both sides, then solve for y.
Solve: dy/dx = xy, y(0) = 2
Separate: dy/y = x dx
Integrate: ln|y| = x²/2 + C
Solve: y = Ae^(x²/2). Use y(0)=2: A=2 → y = 2e^(x²/2)
Model: dP/dt = kP(1 − P/M) where M = carrying capacity. Growth slows as P → M.
Fastest growth occurs when P = M/2 (inflection point of logistic curve).
Euler's Method approximates solutions step by step:
y(new) = y(old) + h · f(x, y)
where h = step size, f(x,y) = dy/dx
dy/dx = x + y, y(0) = 1. Use h = 0.1. Approximate y(0.2).
Step 1: At (0,1): slope = 0+1 = 1. y(0.1) = 1 + 0.1(1) = 1.1
Step 2: At (0.1, 1.1): slope = 0.1+1.1 = 1.2. y(0.2) = 1.1 + 0.1(1.2) = 1.22
Differentiate an equation with respect to time t (not x). Every variable that changes with time needs a rate (·dt).
A 10 ft ladder leans against a wall. The base slides away at 2 ft/s. How fast does the top slide down when the base is 6 ft from the wall?
Relation: x² + y² = 100
Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0
When x=6: y=8. Plug in: 2(6)(2) + 2(8)(dy/dt) = 0 → dy/dt = −3/2 ft/s
Disk/Washer Method (rotating about horizontal axis, integrating with respect to x):
V = π ∫[R(x)² − r(x)²] dx
Shell Method (integrating with respect to x, rotating about y-axis):
V = 2π ∫ x · f(x) dx
Region bounded by y = x² and y = x, rotated about the x-axis.
Intersections: x = 0 and x = 1. On [0,1]: x > x² so outer = x, inner = x²
V = π ∫₀¹ [x² − x⁴] dx = π[x³/3 − x⁵/5]₀¹ = π(1/3 − 1/5) = 2π/15
Visual — Live Animation
Taylor Series Building Up to sin(x)
Watch how more terms of the Taylor series approximate sin(x)
GOOD LUCK ON THE AP EXAM
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