\[ a^m \cdot a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad (a^m)^n = a^{mn} \]

\[ a^0 = 1 \qquad a^{-n} = \frac{1}{a^n} \qquad a^{1/n} = \sqrt[n]{a} \]

\[ \ln(ab) = \ln a + \ln b \qquad \ln\!\left(\frac{a}{b}\right) = \ln a - \ln b \qquad \ln(a^n) = n\ln a \]

\[ \ln(e^x) = x \qquad e^{\ln x} = x \]

\[ \csc x = \frac{1}{\sin x} \qquad \sec x = \frac{1}{\cos x} \qquad \cot x = \frac{1}{\tan x} \]

\[ \tan x = \frac{\sin x}{\cos x} \qquad \cot x = \frac{\cos x}{\sin x} \]

\[ \sin^2 x + \cos^2 x = 1 \qquad 1 + \tan^2 x = \sec^2 x \qquad 1 + \cot^2 x = \csc^2 x \]

\[ \sin 2x = 2\sin x\cos x \]

\[ \cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x = 2\cos^2 x - 1 \]

\[ \sin^2 x = \frac{1 - \cos 2x}{2} \qquad \cos^2 x = \frac{1 + \cos 2x}{2} \]

\[ \sin 0 = 0 \quad \sin\frac{\pi}{6} = \frac{1}{2} \quad \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} \quad \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \quad \sin\frac{\pi}{2} = 1 \]

\[ \cos 0 = 1 \quad \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \quad \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} \quad \cos\frac{\pi}{3} = \frac{1}{2} \quad \cos\frac{\pi}{2} = 0 \]

\[ \tan 0 = 0 \quad \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} \quad \tan\frac{\pi}{4} = 1 \quad \tan\frac{\pi}{3} = \sqrt{3} \quad \tan\frac{\pi}{2} = \text{undef} \]

\(\displaystyle\lim_{x \to c} f(x) = L\) means \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently close to \(c\) (but \(x \neq c\)).

The two-sided limit exists if and only if both one-sided limits exist and are equal: \[\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L\]

If \(g(x) \leq f(x) \leq h(x)\) near \(c\) and \(\displaystyle\lim_{x\to c} g(x) = \lim_{x\to c} h(x) = L\), then \(\displaystyle\lim_{x\to c} f(x) = L\).

\(f\) is continuous at \(c\) if and only if all three conditions hold:

1)   \(f(c)\) is defined

2)   \(\displaystyle\lim_{x \to c} f(x)\) exists

3)   \(\displaystyle\lim_{x \to c} f(x) = f(c)\)

If \(f\) is continuous on \([a, b]\) and \(k\) is any number between \(f(a)\) and \(f(b)\), then there exists at least one \(c\) in \((a, b)\) such that \(f(c) = k\).

If \(\displaystyle\lim_{x\to c}\dfrac{f(x)}{g(x)}\) produces the indeterminate form \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\), then

\[ \lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)} \]

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

\[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \]

\[ \frac{d}{dx}\left[x^n\right] = nx^{n-1} \qquad \frac{d}{dx}\left[c\right] = 0 \qquad \frac{d}{dx}\left[cf(x)\right] = cf'(x) \]

\[ \frac{d}{dx}\left[f \pm g\right] = f' \pm g' \]

\[ \frac{d}{dx}\left[fg\right] = f'g + fg' \]

\[ \frac{d}{dx}\!\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2} \]

\[ \frac{d}{dx}\left[f(g(x))\right] = f'(g(x))\cdot g'(x) \]

\[ \frac{d}{dx}[\sin x] = \cos x \qquad \frac{d}{dx}[\cos x] = -\sin x \qquad \frac{d}{dx}[\tan x] = \sec^2 x \]

\[ \frac{d}{dx}[\cot x] = -\csc^2 x \qquad \frac{d}{dx}[\sec x] = \sec x\tan x \qquad \frac{d}{dx}[\csc x] = -\csc x\cot x \]

\[ \frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}} \qquad \frac{d}{dx}[\arccos x] = \frac{-1}{\sqrt{1-x^2}} \qquad \frac{d}{dx}[\arctan x] = \frac{1}{1+x^2} \]

\[ \frac{d}{dx}\left[e^x\right] = e^x \qquad \frac{d}{dx}\left[a^x\right] = a^x \ln a \]

\[ \frac{d}{dx}[\ln x] = \frac{1}{x} \qquad \frac{d}{dx}[\log_a x] = \frac{1}{x\ln a} \]

If \(g = f^{-1}\), then

\[ g'(x) = \frac{1}{f'(g(x))} \]

Differentiate both sides with respect to \(x\). Whenever differentiating a \(y\)-term, apply the chain rule and multiply by \(\dfrac{dy}{dx}\). Then solve for \(\dfrac{dy}{dx}\).

Take \(\ln\) of both sides, differentiate implicitly, then solve for \(y'\). Especially useful for expressions of the form \(y = f(x)^{g(x)}\).

\[ f(x) \approx f(a) + f'(a)(x - a) \]

\[ e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828\ldots \]

\(c\) is a critical number of \(f\) if \(f'(c) = 0\) or \(f'(c)\) is undefined (with \(f(c)\) defined).

If \(f\) is continuous on \([a, b]\), differentiable on \((a, b)\), and \(f(a) = f(b)\), then there exists at least one \(c\) in \((a, b)\) with \(f'(c) = 0\).

If \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists \(c\) in \((a, b)\) such that

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

At a critical number \(c\) of \(f\):

If \(f'\) changes from \(-\) to \(+\) at \(c\)  →  relative minimum

If \(f'\) changes from \(+\) to \(-\) at \(c\)  →  relative maximum

No sign change  →  neither

If \(f'(c) = 0\):

\(f''(c) > 0\)  →  relative minimum

\(f''(c) < 0\)  →  relative maximum

\(f''(c) = 0\)  →  inconclusive; use First Derivative Test

\(f''(x) > 0\) on \(I\)  →  concave up on \(I\)

\(f''(x) < 0\) on \(I\)  →  concave down on \(I\)

An inflection point occurs at \(x = c\) if \(f''\) changes sign at \(c\).

Write an equation relating two or more quantities. Differentiate both sides with respect to \(t\), applying the chain rule. Substitute known values and solve for the unknown rate.

\[ \int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x \]

If \(f\) is continuous on an interval containing \(a\), then for every \(x\) in that interval,

\[ \frac{d}{dx} \int_a^x f(t)\,dt = f(x) \]

\[ \frac{d}{dx} \int_a^{g(x)} f(t)\,dt = f(g(x))\cdot g'(x) \]

If \(F\) is any antiderivative of \(f\) on \([a, b]\), then

\[ \int_a^b f(x)\,dx = F(b) - F(a) \]

\[ \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) \qquad \int \frac{1}{x}\,dx = \ln|x| + C \]

\[ \int e^x\,dx = e^x + C \qquad \int a^x\,dx = \frac{a^x}{\ln a} + C \]

\[ \int \sin x\,dx = -\cos x + C \qquad \int \cos x\,dx = \sin x + C \]

\[ \int \sec^2 x\,dx = \tan x + C \qquad \int \csc^2 x\,dx = -\cot x + C \]

\[ \int \sec x\tan x\,dx = \sec x + C \qquad \int \csc x\cot x\,dx = -\csc x + C \]

\[ \int \frac{dx}{\sqrt{1-x^2}} = \arcsin x + C \qquad \int \frac{dx}{1+x^2} = \arctan x + C \]

Let \(u = g(x)\), so \(du = g'(x)\,dx\). Rewrite the integral in terms of \(u\), integrate, then back-substitute.

\[ \int f(g(x))\,g'(x)\,dx = \int f(u)\,du \]

\[ \int u\,dv = uv - \int v\,du \]

Replace an infinite bound (or a discontinuity) with a limit:

\[ \int_a^{\infty} f(x)\,dx = \lim_{t\to\infty} \int_a^t f(x)\,dx \]

\[ f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx \]

\[ A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx \]

\[ V = \pi \int_a^b \bigl[f(x)\bigr]^2\,dx \]

\[ V = \pi \int_a^b \Bigl(\bigl[R(x)\bigr]^2 - \bigl[r(x)\bigr]^2\Bigr)\,dx \]

\[ V = \int_a^b A(x)\,dx \]

\[ L = \int_a^b \sqrt{1 + \bigl[f'(x)\bigr]^2}\,dx \]

Position \(s(t)\), velocity \(v(t) = s'(t)\), speed \(= |v(t)|\), acceleration \(a(t) = v'(t)\).

\[ \text{Displacement} = \int_a^b v(t)\,dt \]

\[ \text{Total distance} = \int_a^b |v(t)|\,dt \]

If \(F(x) = \displaystyle\int_a^x f(t)\,dt\), then \(F(x)\) gives the net accumulation of \(f\) from \(a\) to \(x\), and by FTC Part 1, \(F'(x) = f(x)\).

A slope field represents \(\dfrac{dy}{dx} = f(x,y)\) by drawing short line segments with slope \(f(x,y)\) at each point. Solution curves follow the flow of the field.

Starting from \((x_0, y_0)\) with step size \(h\):

\[ y_{n+1} = y_n + h\cdot f(x_n,\, y_n) \]

If \(\dfrac{dy}{dx} = f(x)\,g(y)\), separate and integrate:

\[ \int \frac{dy}{g(y)} = \int f(x)\,dx \]

If \(\dfrac{dy}{dt} = ky\), then

\[ y = y_0\,e^{kt} \]

\[ \frac{dP}{dt} = kP\!\left(1 - \frac{P}{M}\right) \]

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

\[ \frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left[\dfrac{dy}{dx}\right]}{dx/dt} \]

\[ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^{\!2} + \left(\frac{dy}{dt}\right)^{\!2}}\,dt \]

Position: \(\vec{r}(t) = \langle x(t),\; y(t)\rangle\)

Velocity: \(\vec{v}(t) = \langle x'(t),\; y'(t)\rangle\)

Acceleration: \(\vec{a}(t) = \langle x''(t),\; y''(t)\rangle\)

\[ \text{Speed} = |\vec{v}(t)| = \sqrt{\bigl[x'(t)\bigr]^2 + \bigl[y'(t)\bigr]^2} \]

\[ \text{Distance traveled} = \int_a^b |\vec{v}(t)|\,dt \]

\[ x = r\cos\theta \qquad y = r\sin\theta \qquad r^2 = x^2 + y^2 \qquad \tan\theta = \frac{y}{x} \]

\[ \frac{dy}{dx} = \frac{r'(\theta)\sin\theta + r(\theta)\cos\theta}{r'(\theta)\cos\theta - r(\theta)\sin\theta} \]

\[ A = \frac{1}{2}\int_\alpha^\beta \bigl[r(\theta)\bigr]^2\,d\theta \]

\[ A = \frac{1}{2}\int_\alpha^\beta \Bigl(\bigl[r_{\text{outer}}(\theta)\bigr]^2 - \bigl[r_{\text{inner}}(\theta)\bigr]^2\Bigr)\,d\theta \]

A sequence \(\{a_n\}\) converges if \(\displaystyle\lim_{n\to\infty} a_n = L\) for some finite \(L\). Otherwise it diverges.

If \(\displaystyle\lim_{n\to\infty} a_n \neq 0\), then \(\displaystyle\sum a_n\) diverges.

\(\displaystyle\sum_{n=0}^{\infty} ar^n\) converges if \(|r| < 1\) and diverges if \(|r| \geq 1\).

\[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \qquad (|r| < 1) \]

\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \begin{cases} \text{converges} & p > 1 \\ \text{diverges} & p \leq 1 \end{cases} \]

For \(0 \leq a_n \leq b_n\): if \(\sum b_n\) converges then \(\sum a_n\) converges; if \(\sum a_n\) diverges then \(\sum b_n\) diverges.

If \(\displaystyle\lim_{n\to\infty}\dfrac{a_n}{b_n} = L\) where \(0 < L < \infty\), then \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge.

Let \(L = \displaystyle\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|\).

\(L < 1\)  →  converges absolutely   \(L > 1\)  →  diverges   \(L = 1\)  →  inconclusive

\(\displaystyle\sum_{n=1}^\infty (-1)^n a_n\) converges if: (1) \(a_n > 0\), (2) \(a_n\) is decreasing, (3) \(\displaystyle\lim_{n\to\infty} a_n = 0\).

\[ |S - S_n| \leq |a_{n+1}| \]

Absolutely convergent: \(\sum |a_n|\) converges.

Conditionally convergent: \(\sum a_n\) converges but \(\sum |a_n|\) diverges.

A power series \(\displaystyle\sum_{n=0}^\infty c_n(x-a)^n\) has radius of convergence \(R\) found by the Ratio Test. The series converges for \(|x - a| < R\). Always check endpoints separately.

\[ T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(c)}{k!}(x-c)^k \]

\[ e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \qquad \text{(all } x\text{)} \]

\[ \sin x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \qquad \text{(all } x\text{)} \]

\[ \cos x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \qquad \text{(all } x\text{)} \]

\[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots \qquad (|x| < 1) \]

\[ \ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \qquad (-1 < x \leq 1) \]

\[ \arctan x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \qquad (|x| \leq 1) \]

\[ \left|R_n(x)\right| \leq \frac{M}{(n+1)!}\,|x - c|^{\,n+1} \]