Q5 (4.3 pts)
Ch 2.4 — The Precise Definition of a Limit (p.113)
The ε-δ definition of a limit:
\[\lim_{x\to a}f(x) = L \iff \forall\varepsilon>0,\;\exists\delta>0: 0<|x-a|<\delta \implies |f(x)-L|<\varepsilon\]
Intuition: "If \(x\) is close enough to \(a\) (within \(\delta\)), then \(f(x)\) is as close as we want to \(L\) (within \(\varepsilon\))."
Exam format: Given \(\varepsilon\), find the maximum \(\delta\).
Strategy:
1. Solve the inequality \(|f(x)-L|<\varepsilon\) to find the range of \(x\)
2. Compare with \(|x-a|<\delta\) to get candidate values for \(\delta\)
3. Take the minimum of the two candidates as the maximum \(\delta\)
Find the maximum \(\delta\) such that if \(|x-2|<\delta\) then \(|x^2-4|<1\).
Step 1 · Ch 2.4: Solve the conclusion inequality
\(|x^2-4|<1 \Rightarrow -1
Near \(x=2\) (so \(x>0\)): \(\sqrt{3}
Step 2 · Ch 2.4: Compute δ candidates
From \(|x-2|<\delta\): \(2-\delta
Left bound: \(2-\delta=\sqrt{3} \Rightarrow \delta=2-\sqrt{3}\approx 0.268\)
Right bound: \(2+\delta=\sqrt{5} \Rightarrow \delta=\sqrt{5}-2\approx 0.236\)
Step 3 · Ch 2.4: Maximum δ = min of candidates
\(\delta = \min(2-\sqrt{3},\;\sqrt{5}-2) = \sqrt{5}-2\)
Answer: (2) \(\sqrt{5}-2\)
Q6 (4.3 pts)
Ch 2.6 — Limits at Infinity
Ch 3.1 — The Natural Exponential Function
Limits of the natural exponential function \(e^x\):
\[\lim_{x\to\infty}e^x = \infty, \quad \lim_{x\to-\infty}e^x = 0\]
For composite forms, evaluate the exponent's limit first, then substitute.
One-sided limits to watch:
\[\lim_{x\to 0^+}\frac{1}{x}=+\infty, \quad \lim_{x\to 0^-}\frac{1}{x}=-\infty\]
\(\displaystyle\lim_{x \to 0^+} e^{-1/x}\)
Step 1 · Ch 2.2: One-sided limit
As \(x\to 0^+\): \(\frac{1}{x}\to +\infty\)
Step 2 · Sign handling
Therefore \(-\frac{1}{x}\to -\infty\)
Step 3 · Ch 3.1: Exponential property
\(e^{-\infty}=0\)
Answer: (4) 0
Q7 (4.4 pts)
Ch 3.1 — Derivatives of Polynomials and Exponential Functions (p.183)
Power Rule:
\[\frac{d}{dx}x^n = nx^{n-1}\]
This applies to positive integers, negative integers, fractions, and all real exponents.
Key conversions:
\(\sqrt{x} = x^{1/2}\), \(\frac{1}{x} = x^{-1}\), \(\frac{1}{\sqrt{x}} = x^{-1/2}\), \(\frac{1}{x\sqrt{x}} = x^{-3/2}\)
Find \(f'(x)\) when \(f(x)=\dfrac{2}{\sqrt{x}}\).
Step 1 · Exponent conversion
\(f(x) = 2x^{-1/2}\)
Step 2 · Ch 3.1: Power Rule
\(f'(x) = 2\cdot\left(-\frac{1}{2}\right)x^{-1/2-1} = -x^{-3/2}\)
Step 3 · Restore original form
\(f'(x) = -\dfrac{1}{x^{3/2}} = -\dfrac{1}{x\sqrt{x}}\)
Answer: (1) \(-\frac{1}{x\sqrt{x}}\)
Q8 (4.4 pts)
Ch 3.1 — Exponential Derivative
Ch 3.4 — The Chain Rule (basic)
Derivative of the natural exponential:
\[\frac{d}{dx}e^x = e^x\]
Combined with Chain Rule:
\[\frac{d}{dx}e^{kx} = ke^{kx}\]
Essential exponent laws:
\(e^{\ln a} = a\), \(e^{k\ln a} = e^{\ln a^k} = a^k\)
Find \(f'(\ln 2)\) when \(f(x)=2e^{2x}\).
Step 1 · Ch 3.1 + 3.4: Exponential differentiation
\(f'(x) = 2\cdot 2e^{2x} = 4e^{2x}\)
Step 2 · Exponent law: \(e^{k\ln a}=a^k\)
\(f'(\ln 2) = 4e^{2\ln 2} = 4\cdot e^{\ln 2^2} = 4\cdot e^{\ln 4}\)
Step 3 · Ch 3.1: \(e^{\ln a}=a\)
\(= 4\cdot 4 = 16\)
Answer: (3) 16
Q9 (4.4 pts)
Ch 3.2 — The Product and Quotient Rules (p.190)
Product Rule:
\[(fg)' = f'g + fg'\]
In words: "(derivative of first) × (second) + (first) × (derivative of second)"
Common mistake: \((fg)' \neq f'\cdot g'\)
Find \(f'(\ln 2)\) when \(f(x)=xe^{2x}\).
Step 1 · Ch 3.2: Apply Product Rule
Let \(u=x\), \(v=e^{2x}\). Then \(u'=1\), \(v'=2e^{2x}\) (by Ch 3.4).
Step 2 · Product Rule result
\(f'(x) = 1\cdot e^{2x} + x\cdot 2e^{2x} = e^{2x}(1+2x)\)
Step 3 · Substitution + exponent calculation
\(f'(\ln 2) = e^{2\ln 2}(1+2\ln 2) = 4(1+2\ln 2) = 4+8\ln 2\)
Answer: (5) \(4+8\ln 2\)
Q10 (4.5 pts)
Ch 3.5 — Implicit Differentiation (p.213)
Implicit Differentiation: When \(y\) is not given as an explicit function of \(x\), and the equation has the form \(F(x,y)=0\):
Method:
1. Differentiate both sides with respect to \(x\)
2. Every time you differentiate a \(y\)-term, multiply by \(\frac{dy}{dx}\) (= \(y'\)) — this is the Chain Rule
3. Solve for \(y'\)
Why Chain Rule applies to \(y\)-terms:
\[\frac{d}{dx}y^2 = \frac{d}{dy}(y^2)\cdot\frac{dy}{dx} = 2y\cdot y'\]
Find the slope of the tangent line to \(x^2+2xy+2y^2=5\) at \((1,1)\).
Step 1 · Ch 3.5: Differentiate both sides
\(\frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) + \frac{d}{dx}(2y^2) = 0\)
Step 2 · Ch 3.1: Power Rule
\(\frac{d}{dx}(x^2) = 2x\)
Step 3 · Ch 3.2: Product Rule for \(2xy\)
\(\frac{d}{dx}(2xy) = 2y + 2xy'\)
Step 4 · Ch 3.5: Chain Rule for \(y^2\)
\(\frac{d}{dx}(2y^2) = 4yy'\)
Step 5 · Combine and simplify
\(2x + 2y + 2xy' + 4yy' = 0\)
Step 6 · Substitute (1,1)
\(2+2+2y'+4y'=0 \Rightarrow 4+6y'=0 \Rightarrow y'=-\frac{2}{3}\)
Answer: (2) \(-\frac{2}{3}\)
Q11 (4.5 pts)
Ch 2.2 — The Limit of a Function (p.89) — Infinite Limits
Infinite Limits: \(\lim_{x\to a}f(x) = \infty\) requires \(f(x)\) to diverge to positive infinity from both sides.
Key distinction:
• \(\lim_{x\to 0}\frac{1}{x}\): left limit \(= -\infty\), right limit \(= +\infty\) → DNE (cannot write "\(=\infty\)")
• \(\lim_{x\to 0}\frac{1}{x^2}\): left limit \(= +\infty\), right limit \(= +\infty\) → can write "\(=\infty\)"
Reason: \(1/x\) changes sign, but \(1/x^2\) is always positive.
Which are true?
Ā \(\lim_{x\to 0}x=0\),
&Bmacr; \(\lim_{x\to 0}\frac{1}{x}=\infty\),
&Cmacr; \(\lim_{x\to 0}\frac{1}{x^2}=\infty\)
Step 1 · Ch 2.2: Continuous function limit
A: \(\lim_{x\to 0}x=0\) ✓ TRUE
Step 2 · Ch 2.2: Compare one-sided limits
B: Left limit \(=-\infty\), right limit \(=+\infty\). They do not agree ✗ FALSE
Step 3 · Ch 2.2: Even power
C: \(x^2>0\), so both sides → \(+\infty\) ✓ TRUE
Answer: (3) A and C only
Q12 (4.5 pts)
Ch 2.6 — Limits at Infinity (p.135)
Application of the rationalization technique from Q4 to true/false judgments. (Same concepts as Q4.)
Determine which of A, B, C are true.
Step 1 · Rationalization
A: \(\frac{2}{\sqrt{x^2+2}+x}\to\frac{2}{\infty}=0\) ✓ TRUE
Step 2 · Rationalization + quick formula
B: \(\frac{6x}{\sqrt{4x^2+6x}+2x}\to\frac{6}{2+2}=\frac{3}{2}\neq 3\) ✗ FALSE
Step 3 · Approximation
C: \(\sqrt{4x^2+\cdots}\approx 2x\), then \(2x-3x=-x\to-\infty\neq 2\) ✗ FALSE
Answer: (1) A only
Q13 (4.6 pts)
Ch 2.7 — Derivatives and Rates of Change (tangent line, p.148)
Equation of the tangent line at \((a, f(a))\):
\[y - f(a) = f'(a)(x - a)\]
To find the y-intercept, substitute \(x=0\).
Find the y-intercept of the tangent to \(y=-x^2\) at \(x=2\).
Step 1 · Function value
\(f(2)=-4\)
Step 2 · Ch 3.1: Power Rule
\(f'(x)=-2x\), so \(f'(2)=-4\)
Step 3 · Ch 2.7: Tangent line formula
\(y-(-4)=-4(x-2) \Rightarrow y=-4x+4\)
Step 4 · y-intercept
\(x=0\): \(y=4\)
Answer: (2) 4
Q14 (4.6 pts)
Ch 3.5 — Implicit Differentiation (tangent to implicit curve)
Find the y-intercept of the tangent to \(xy=8\) at \(x=2\).
Step 1 · Function value
At \(x=2\): \(y=4\)
Step 2 · Ch 3.5: Implicit differentiation
\(y+xy'=0 \Rightarrow y'=-y/x\)
Step 3 · Substitution
\(y'(2)=-4/2=-2\)
Step 4 · Tangent line
\(y-4=-2(x-2) \Rightarrow y=-2x+8\). y-intercept \(=8\)
Answer: (3) 8
Q15 (4.6 pts)
Ch 3.5 — Implicit Differentiation
Ch 3.1 — Power Rule (fractional exponent)
Fractional exponent differentiation in implicit context:
\[\frac{d}{dx}x^{2/3} = \frac{2}{3}x^{-1/3}\]
Find the slope of the tangent to \(x^{2/3}+y^{2/3}=13\) at \((8,27)\).
Step 1 · Verification
\(8^{2/3}+27^{2/3}=(\sqrt[3]{8})^2+(\sqrt[3]{27})^2=4+9=13\) ✓
Step 2 · Ch 3.5 + 3.1: Implicit differentiation
\(\frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}y'=0\)
Step 3 · Solve for \(y'\)
\(y'=-\dfrac{x^{-1/3}}{y^{-1/3}}=-\left(\dfrac{y}{x}\right)^{1/3}\)
Step 4 · Substitution
\(y'=-\left(\dfrac{27}{8}\right)^{1/3}=-\dfrac{3}{2}\)
Answer: (1) \(-\frac{3}{2}\)
Q16 (4.7 pts)
Ch 3.6 — Derivatives of Logarithmic and Inverse Trigonometric Functions (p.223)
Derivation of the derivative of arcsin:
\(y=\arcsin x \Rightarrow \sin y = x\)
Differentiate both sides: \(\cos y\cdot y'=1 \Rightarrow y'=\frac{1}{\cos y}\)
Since \(\cos y = \sqrt{1-\sin^2 y}=\sqrt{1-x^2}\) (because \(-\pi/2\le y\le\pi/2\) means \(\cos y\ge 0\)):
Final formula:
\[\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}\]
\(\frac{d}{dx}(\arcsin x)\) at \(x=\frac{4}{5}\)
Step 1 · Ch 3.6: arcsin derivative formula
\(\frac{1}{\sqrt{1-x^2}}\)
Step 2 · Substitution
\(\frac{1}{\sqrt{1-(4/5)^2}}=\frac{1}{\sqrt{1-16/25}}=\frac{1}{\sqrt{9/25}}\)
Step 3 · Simplify the root
\(=\frac{1}{3/5}=\frac{5}{3}\)
Answer: (4) \(\frac{5}{3}\)
Q17 (4.7 pts)
Ch 3.4 — The Chain Rule (p.203)
Chain Rule: For composite functions:
\[\frac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)\]
Remember: "derivative of the outer × derivative of the inner"
Exponential composition:
\[\frac{d}{dx}e^{g(x)} = e^{g(x)}\cdot g'(x)\]
\(\frac{d}{dx}\{e^{(x^2)}\}\) at \(x=2\)
Step 1 · Ch 3.4: Identify Chain Rule
Outer: \(e^u\), Inner: \(u=x^2\)
Step 2 · Outer derivative × Inner derivative
\(e^{x^2}\cdot 2x\)
Step 3 · Substitution
\(e^{2^2}\cdot 2\cdot 2 = 4e^4\)
Answer: (5) \(4e^4\)
Q18 (4.8 pts)
Ch 3.4 — The Chain Rule (double composition)
Double Chain Rule: Applying the Chain Rule twice in succession:
\[\frac{d}{dx}e^{e^x} = e^{e^x}\cdot\frac{d}{dx}(e^x) = e^{e^x}\cdot e^x\]
\(\frac{d}{dx}\{e^{(e^x)}\}\) at \(x=\ln 2\)
Step 1 · Ch 3.4: First Chain Rule
Outer: \(e^u\), Inner: \(u=e^x\). Result: \(e^{e^x}\cdot e^x\)
Step 2 · Substitute: \(e^{\ln 2}=2\)
\(e^x = e^{\ln 2}=2\), so \(e^{e^x}=e^2\)
Step 3 · Calculate
\(e^2\cdot 2 = 2e^2\)
Answer: (2) \(2e^2\)
Q19 (4.8 pts)
Ch 3.2 — Product Rule
Ch 3.3 — Derivatives of Trigonometric Functions
This problem combines three concepts:
1. Product Rule: \((e^x\cos x)' = e^x\cos x - e^x\sin x = e^x(\cos x - \sin x)\)
2. Trig derivative: \((\cos x)' = -\sin x\)
3. Trig ratios from arctan: If \(\theta=\arctan(3/4)\), then from the 3-4-5 right triangle:
opposite = 3, adjacent = 4, hypotenuse = 5
\(\sin\theta=3/5\), \(\cos\theta=4/5\)
\(\frac{d}{dx}(e^x\cos x)\) at \(x=\arctan(3/4)\)
Step 1 · Ch 3.2: Product Rule
\(\frac{d}{dx}(e^x\cos x) = e^x\cos x + e^x(-\sin x) = e^x(\cos x-\sin x)\)
Step 2 · Trig ratio calculation
\(x=\arctan(3/4)\): from the 3-4-5 right triangle, \(\sin x=\frac{3}{5}\), \(\cos x=\frac{4}{5}\)
Step 3 · Substitution
\(e^{\arctan(3/4)}\left(\frac{4}{5}-\frac{3}{5}\right)=\frac{1}{5}e^{\arctan(3/4)}\)
Answer: (5) \(\frac{1}{5}e^{\arctan(3/4)}\)
Q20 (4.9 pts)
Ch 3.4 — Chain Rule
Ch 3.3 — Trigonometric Functions
\(\frac{d}{dx}(e^{\sin x})\) at \(x=\arctan(3/4)\)
Step 1 · Ch 3.4: Chain Rule
\(\frac{d}{dx}e^{\sin x} = e^{\sin x}\cdot\cos x\)
Step 2 · Trig ratios
\(\sin(\arctan(3/4))=\frac{3}{5}\), \(\cos(\arctan(3/4))=\frac{4}{5}\)
Step 3 · Substitution
\(e^{3/5}\cdot\frac{4}{5}=\frac{4}{5}e^{3/5}\)
Answer: (3) \(\frac{4}{5}e^{3/5}\)
Q21 (5.0 pts)
Ch 3.6 — Inverse Trigonometric Functions (derivation via implicit differentiation)
The derivation process of the arctan derivative itself is tested. Key steps:
1. From \(y=\arctan 2x\), use the inverse function definition: \(x=\frac{1}{2}\tan y\)
2. Differentiate both sides w.r.t. \(y\): \(\frac{dx}{dy}=\frac{1}{2}\sec^2 y\)
3. Trig identity: \(\sec^2 y = 1+\tan^2 y = 1+(2x)^2 = 1+4x^2\)
4. Substitute: \(\frac{dx}{dy}=\frac{1+4x^2}{2}\)
Choose the correct expression for A (\(\frac{dx}{dy}\)) where \(y=\arctan 2x\).
Step 1 · Ch 3.6: Inverse function relation
\(x=\frac{1}{2}\tan y\)
Step 2 · Differentiate w.r.t. \(y\)
\(\frac{dx}{dy}=\frac{1}{2}\sec^2 y\)
Step 3 · Trig identity: \(\sec^2=1+\tan^2\)
\(\sec^2 y=1+\tan^2 y=1+(2x)^2=1+4x^2\)
Step 4 · Substitution
\(\frac{dx}{dy}=\frac{1+4x^2}{2}\)
Answer: (4) \(\frac{1+4x^2}{2}\)
Q22 (5.1 pts) — Highest Point Value
Ch 9.3 — Separable Equations (Orthogonal Trajectories, p.619)
Orthogonal Trajectories: A family of curves that intersects a given family of curves at right angles at every intersection point.
Two curves are orthogonal ⇔ the product of slopes at the intersection is \(-1\):
\[m_1 \cdot m_2 = -1 \quad \Rightarrow \quad m_2 = -\frac{1}{m_1}\]
4-step method:
1. Eliminate the constant \(C\) from the given family by differentiating
2. Find \(y'\)
3. Orthogonal condition: replace with \(y'_\perp = -1/y'\)
4. Solve the new ODE by separation of variables
Find the orthogonal trajectories of \(\frac{x^2}{4}+y^2=C\).
Step 1 · Differentiate to eliminate C
\(\frac{x}{2}+2yy'=0 \Rightarrow y'=-\frac{x}{4y}\)
Step 2 · Apply orthogonal condition
\(y'_\perp = -\frac{1}{-x/(4y)} = \frac{4y}{x}\)
Step 3 · Ch 9.3: Separation of variables
\(\frac{dy}{y}=\frac{4\,dx}{x}\)
Step 4 · Integrate
\(\ln|y|=4\ln|x|+C_1 \Rightarrow y=Kx^4\)
Answer: (1) \(y=Kx^4\)
Final Checklist — Stewart's Calculus 9/E
- Ch 2.7 — Can you instantly recognize derivative definitions disguised as limit problems?
- Ch 2.4 — Can you solve ε-δ problems (solve inequality, then take min)?
- Ch 2.6 — Do you remember the rationalization quick formula \(\frac{b}{2\sqrt{a}}\)?
- Ch 3.1 — Can you apply the Power Rule to negative and fractional exponents?
- Ch 3.1 — Are you fluent with \(e^{k\ln a}=a^k\)?
- Ch 3.2 — Do you apply the Product Rule correctly? (Common mistake: \((fg)'=f'g'\))
- Ch 3.4 — Can you handle double-composition Chain Rule?
- Ch 3.5 — Do you correctly isolate \(y'\) in implicit differentiation?
- Ch 3.6 — Do you know the arcsin and arctan derivative formulas and their derivations?
- Trig ratios — Does arctan(3/4) immediately bring to mind the 3-4-5 triangle?
- Ch 9.3 — Can you recall all 4 steps for orthogonal trajectories in order?