MATH 100 & 180: What to Memorize Before the Exam

Only the essentials: the things you can't work out quickly in an exam. In this course log⁡\log means the natural log, ln⁡\ln. Practice them as flashcards.

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Limits at Infinity & Dominant Terms

  • Who wins as x→∞\text{Who wins as } x \to \infty
    log⁡x≪xp≪ex\log x \ll x^p \ll e^x
    Any positive power beats log⁡x\log x; exe^x beats any power. This is for terms added or subtracted: multiplied or divided pieces all stay.
  • lim⁡x→−∞ex=0\lim_{x \to -\infty} e^x = 0
    So as x→−∞x \to -\infty the polynomial terms win.
  • lim⁡x→−∞e−x=+∞\lim_{x \to -\infty} e^{-x} = +\infty
    −x→+∞-x \to +\infty, so e−xe^{-x} grows.
  • lim⁡x→−∞xn={+∞n even−∞n odd\lim_{x \to -\infty} x^n = \begin{cases} +\infty & n \text{ even} \\ -\infty & n \text{ odd} \end{cases}
    Odd powers keep the minus sign: x3→−∞x^3 \to -\infty, x2→+∞x^2 \to +\infty.
  • lim⁡x→±∞cxn=0\lim_{x \to \pm\infty} \dfrac{c}{x^n} = 0
    For any n>0n > 0, at both ends.
  • lim⁡x→−∞ax={0a>1+∞0<a<1\lim_{x \to -\infty} a^x = \begin{cases} 0 & a > 1 \\ +\infty & 0 < a < 1 \end{cases}
    So 2x→02^x \to 0 but (0.5)x→∞(0.5)^x \to \infty: a base below 11 flips.
  • ∣x∣={xif x≥0−xif x<0\lvert x \rvert = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}
    Negative inside: flip the sign. So ∣−3∣=3\lvert -3 \rvert = 3, and ∣x−2∣=2−x\lvert x - 2 \rvert = 2 - x when x<2x < 2.
  • x2=∣x∣\sqrt{x^2} = |x|
    Equals −x-x when x→−∞x \to -\infty.

The Derivative from the Definition

  • f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0}\dfrac{f(x + h) - f(x)}{h}
    When the question says “use the definition”.

Differentiation Rules

  • ddx xn=n xn−1\dfrac{d}{dx}\,x^n = n\,x^{n-1}
    Any constant nn: roots and fractions too. Rewrite x=x1/2\sqrt{x} = x^{1/2}, 1x3=x−3\dfrac{1}{x^3} = x^{-3} first.
  • ddx ex=ex\dfrac{d}{dx}\,e^x = e^x
  • ddx ax=axlog⁡a\dfrac{d}{dx}\,a^x = a^x \log a
    For a constant base like 2x2^x. Not x ax−1x\,a^{x-1}.
  • ddx log⁡x=1x\dfrac{d}{dx}\,\log x = \dfrac{1}{x}
    log⁡\log is the natural log. With an inside: u′u\dfrac{u'}{u}.
  • ddx sin⁡x=cos⁡x\dfrac{d}{dx}\,\sin x = \cos x
  • ddx cos⁡x=−sin⁡x\dfrac{d}{dx}\,\cos x = -\sin x
    Watch the minus sign.
  • ddx tan⁡x=sec⁡2x\dfrac{d}{dx}\,\tan x = \sec^2 x
  • (fg)′=f′g+fg′(fg)' = f'g + fg'
  • (fg)′=f′g−fg′g2\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}
    "Low d-high minus high d-low, over low squared."
  • ddx f(g(x))=f′(g(x))⋅g′(x)\dfrac{d}{dx}\,f(g(x)) = f'(g(x))\cdot g'(x)
    Outside derivative, inside kept, times inside derivative.

Tangent Lines & Linear Approximation

  • Tangent line at x=a\text{Tangent line at } x = a
    y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x - a)
    Point-slope form: a point and a slope. The derivative f′(a)f'(a) is only the slope, not the line.
  • Linear approximation at x=a\text{Linear approximation at } x = a
    L(x)=f(a)+f′(a)(x−a)L(x) = f(a) + f'(a)(x - a)
    Exactly the tangent line at x=ax = a, solved for yy.

Inverse Trig Functions

  • ddx arcsin⁡x=11−x2\dfrac{d}{dx}\,\arcsin x = \dfrac{1}{\sqrt{1 - x^2}}
  • ddx arctan⁡x=11+x2\dfrac{d}{dx}\,\arctan x = \dfrac{1}{1 + x^2}

Curve Sketching

  • When f′(x)>0\text{When } f'(x) > 0
    f is increasingf \text{ is increasing}
    The graph goes up, left to right.
  • When f′(x)<0\text{When } f'(x) < 0
    f is decreasingf \text{ is decreasing}
    The graph goes down, left to right.
  • When f′′(x)>0\text{When } f''(x) > 0
    f is concave up (∪)f \text{ is concave up } (\cup)
    The slope f′f' is increasing.
  • When f′′(x)<0\text{When } f''(x) < 0
    f is concave down (∩)f \text{ is concave down } (\cap)
    The slope f′f' is decreasing.

Optimization

  • Right triangle\text{Right triangle}
    a2+b2=c2a^2 + b^2 = c^2
  • Circle: area, circumference\text{Circle: area, circumference}
    πr2,  2πr\pi r^2,\ \ 2\pi r
  • Sphere volume\text{Sphere volume}
    43πr3\tfrac{4}{3}\pi r^3
  • Cylinder volume\text{Cylinder volume}
    πr2h\pi r^2 h
    Closed surface area 2πr2+2πrh2\pi r^2 + 2\pi r h.
  • Cone volume\text{Cone volume}
    13πr2h\tfrac{1}{3}\pi r^2 h
    Use similar triangles to remove rr or hh first.

Taylor Polynomials & Newton's Method

  • Taylor polynomial about x=a\text{Taylor polynomial about } x = a
    Tn(x)=∑k=0nf(k)(a)k!(x−a)kT_n(x) = \sum_{k=0}^{n} \dfrac{f^{(k)}(a)}{k!}(x - a)^k
  • ex=1+x+x22+x36+⋯e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots
  • sin⁡x=x−x36+⋯\sin x = x - \dfrac{x^3}{6} + \cdots
  • cos⁡x=1−x22+x424−⋯\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots
  • log⁡(1+x)=x−x22+x33−⋯\log(1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots

Used in every topic

Log and exponent rules

Natural log only. You need these for log differentiation and limits.
  • log⁡(ab)=log⁡a+log⁡b\log(ab) = \log a + \log b
  • log⁡ab=log⁡a−log⁡b\log\dfrac{a}{b} = \log a - \log b
  • log⁡(ak)=klog⁡a\log(a^k) = k\log a
  • elog⁡x  and  log⁡(ex)=xe^{\log x}\ \text{ and }\ \log(e^x) = x
    log⁡\log and exe^x undo each other. The first needs x>0x > 0.
  • log⁡1=0\log 1 = 0
    So log⁡x=0\log x = 0 exactly when x=1x = 1, and log⁡e=1\log e = 1
  • Get rid of a log: use e\text{Get rid of a log: use } e
    log⁡x=c  ⟹  x=ec\log x = c \;\Longrightarrow\; x = e^{c}
    Raise ee to the power of both sides, because elog⁡x=xe^{\log x} = x
  • Get rid of e: use log\text{Get rid of } e\text{: use log}
    ex=c  ⟹  x=log⁡ce^{x} = c \;\Longrightarrow\; x = \log c
    Take the log of both sides, because log⁡(ex)=x\log(e^{x}) = x (needs c>0c > 0)

Trig

The definitions, values and identities that come up every term.
  • tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{\sin x}{\cos x}
  • sec⁡x=1cos⁡x\sec x = \dfrac{1}{\cos x}
  • sin at 0, π4, π2, 3π4, π\text{sin at } 0,\ \tfrac{\pi}{4},\ \tfrac{\pi}{2},\ \tfrac{3\pi}{4},\ \pi
    0, 22, 1, 22, 00,\ \tfrac{\sqrt{2}}{2},\ 1,\ \tfrac{\sqrt{2}}{2},\ 0
  • cos at 0, π4, π2, 3π4, π\text{cos at } 0,\ \tfrac{\pi}{4},\ \tfrac{\pi}{2},\ \tfrac{3\pi}{4},\ \pi
    1, 22, 0, −22, −11,\ \tfrac{\sqrt{2}}{2},\ 0,\ -\tfrac{\sqrt{2}}{2},\ -1
    And tan⁡π4=1\tan\tfrac{\pi}{4} = 1.
  • sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1
  • 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x
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