Table of Contents
$$\begin{align}&\sin{x}=x+\mathcal{O}(x),\\&\cos{x}=1-\frac{1}{2}x^2+\mathcal{O}(x^2),\\&a^x=1+x\ln{a}+\mathcal{O}(x)(a\gt 0),\\&\ln{(1+x)}=x+\mathcal{O}(x),\\&e^x=1+x+\mathcal{O}(x),\\&(1+x)^a=1+ax+\mathcal{O}(x),\\&\tan{x}=x+\mathcal{O}(x),\\&\sinh{x}=x+\mathcal{O}(x),\\&\cosh{x}=1+\frac{x^2}{2}+\mathcal{O}(x^2),\\&\tanh{x}=x+\mathcal{O}(x)\end{align}$$
$$\begin{array}{l | l} \textbf{Easier}&\textbf{Harder}\\ (x^\alpha)'=\alpha x^{\alpha-1}, \forall\alpha & (\arcsin x)' = \frac{1}{\sqrt{1-x^2}},-1\lt x\lt 1\\ (\sin{x})’=\cos{x} & (\arccos{x})'=-\frac{1}{\sqrt{1-x^2}},(-1\lt x\lt 1)\\ (\cos{x})’=-\sin{x} & (\arctan{x})'=\frac{1}{1+x^2}\\ (\log_a{x})’=\frac{1}{x\ln{a}} & (\textrm{arccot}\;x)'=-\frac{1}{1+x^2}\\ (\ln{x})’=\frac{1}{x}(x\gt 0) & (\sinh{x})’=\cosh{x}\\ (a^x)’=a^x\ln{a} & (\cosh{x})’=\sinh{x}\\ (e^x)’=e^x & (\tanh{x})’=\frac{1}{(\cosh{x})^2}\\ (\tan{x})’= \sec^2x,(x\neq \frac{\pi}{2}+\pi n,n\in \mathbb{Z}) & (\coth{x})’=-\frac{1}{(\sinh{x})^2}(x\neq 0)\\ (\cot{x})’=-\csc^2x,(x\neq\pi n,n\in \mathbb{Z}) & \end{array}$$
If $f(x)\leq g(x)\leq h(x)$ and $\lim_{x\to 0}f(x)=L$ and $\lim_{x\to 0}h(x)=L$, we can deduce $\lim_{x\to 0}g(x)=L$
If a limit produces indeterminacies: $$ \lim_{x\to a}\frac{\alpha(x)}{\beta(x)}= \lim_{x\to a}\frac{\alpha'(x)}{\beta'(x)} $$ It will work for the following indeterminacies: $$ \frac{0}{0},\;\;\frac{\infty}{\infty},\;\;0\times\infty,\;\;1^\infty,\;\;0^0,\;\;\infty^0,\;\;\infty-\infty $$
To manipulate the limit $\lim_{r \to \infty} \left(1 + \frac{1}{r^3}\right)^{3r^3}$ to use the special limit $\lim_{r \to \infty} \left(1 + \frac{1}{r}\right)^r = e$, we proceed as follows: Start with the original limit: $$ \lim_{r \to \infty} \left( 1 + \frac{1}{r^3} \right)^{3r^3} $$ Next, let $x = r^3$. As $r \to \infty$, $x \to \infty$ as well. Rewrite the expression in terms of $x$: $$ \left( 1 + \frac{1}{x} \right)^{3x} $$ Notice that the exponent $3r^3$ becomes $3x$. This can be expressed as: $$ \left[ \left( 1 + \frac{1}{x} \right)^x \right]^3 $$ We know from the special limit that: $$ \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right)^x = e $$ Therefore, taking the limit as $x \to \infty$: $$ \left[ \left( 1 + \frac{1}{x} \right)^x \right]^3 \to e^3 $$ Thus, the limit simplifies to: $$ \lim_{r \to \infty} \left( 1 + \frac{1}{r^3} \right)^{3r^3} = e^3 $$ In conclusion, we have: $$ \boxed{e^3} $$ Personal note: this this trick means that we can use the identity with: $$ \lim_{r\to\infty}\left(1+x\right)^\frac{1}{x} $$
$$ (uv)^{(n)}=\sum_{i=0}^nc^i_nu^{(i)}v^{(n-i)},\;\;\; c^i_n=\frac{n!}{i!(n-i)!} $$
$$ \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}(\frac{dy/dt}{dx/dt})}{\frac{dx}{dt}} $$
$$ (f^{-1}(y_0))'=\frac{1}{f'(x_0)}\implies (f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))} $$
For $z=f(u(x,y),v(x,y))$ $$ \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y} $$
If a function $z=f(x,y)$ is differentiable at point $M_0(x_0,y_0$) of the function domain, then at point $N_0(x_0,y_0,f(x_0,y_0))$ there is a tangent plane to the surface $S$ defined by: $$ \frac{\partial z}{\partial x}(M_0)(x-x_0)+\frac{\partial z}{\partial y}(M_0)(y-y_0)-(z-f(x_0,y_0))=0 $$ Or for function $F(x,y,z)$: $$ \frac{\partial F}{\partial x} (x-x_0) + \frac{\partial F}{\partial y} (y-y_0) + \frac{\partial F}{\partial z} (z-z_0)=0 $$
For a differentiable function $f(x_1,x_2,...,x_m),$ we define the differentials $dx_1,dx_2,...,dx_m$ to be independent variables, that is, they can be given any values. Then the differential $df$, also called the total differential, is defined by $$ df=\frac{\partial f}{\partial x_1}dx_1+\frac{\partial f}{\partial x_2}dx_2+...+\frac{\partial f}{\partial x_m}dx_m $$
For $u=f(M)$ at a point $M_0(x_1^0,...,x^0_m)$ and partial derivatives with respect to $x_k$, there is a stationary point if: $$ \frac{\partial u}{\partial x_k}(M_0)=0 $$ Furthermore, the following expression yields more information about the function. $$ R=\frac{\partial^2 u}{\partial x^2}(M_0)\cdot\frac{\partial^2 u}{\partial y^2}(M_0)-(\frac{\partial^2u}{\partial x\partial y}(M_0))^2 $$
If $F(x,y)=0$ defines y implicitly as a function of x, then $$ \frac{dy}{dx}=-\frac{F_x}{F_y}, \textrm{ if } F_y\neq0 $$ $$ f''(x)=\frac{d}{dx}\begin{bmatrix}-\frac{F_x(x,f(x))}{F_y(x,f(x))}\end{bmatrix} $$
The most fun. Related to cartesian coordinates: $$ x=r\cos{\theta},\;\;\;\;\;y=r\sin{\theta} $$
Polar, along with a height, $z$. $$ x=r\cos{\theta},\;\;\;\;\;y=r\sin{\theta}\;\;\;\;\;z=z $$
Two sets of angles and a radius. $$ x=r\cos{\theta}\sin{\phi},\;\;\;\;\;y=r\sin{\theta}\sin{\phi}\;\;\;\;\;z=r\cos{\phi} $$
The one you’re probably going to need more. $$ \mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos{\phi} $$ Magnitude $|\mathbf{a}|$: $$|\mathbf{a}|=\sqrt{\mathbf{a}\cdot \mathbf{a}}$$ Components: $$ (Q\mathbf{\hat{i}}+R\mathbf{\hat{j}})\cdot(S\mathbf{\hat{i}}+T\mathbf{\hat{j}})=QS+TR $$
More complicated, and orthogonal. Use right hand rule: CAB Note: For vectors, $a\times b = -b\times a$ and $a^2=0$. $$ \mathbf{c}=\mathbf{a}\times \mathbf{b}\implies|\mathbf{c}|=|\mathbf{a}||\mathbf{b}|\sin{\phi} $$ Components: $$ \mathbf{a}\times \mathbf{b}=(a_2b_3-a_3b_2)i-(a_3b_1-a_1b_3)j+(a_1b_2-a_2b_1)k $$ Is possible to work it out by expanding the brackets and remembering $i^2,j^2,k^2=0$, and the vector circle. i→j→k→i.
A unit vector exists for any non-zero vector a and has the length equal to 1. In Cartesian Coordinates, $a=(P(x,y,z), Q(x,y,z),R(x,y,z))$. $u=\frac{1}{\sqrt{P^2+Q^2+R^2}}(P,Q,R) = (\cos\alpha, \cos\beta, \cos\gamma)$, where $\alpha,\beta,\gamma$ are the angles between vector a and the corresponding axes. $$ \nabla=\frac{\partial}{\partial x}i+\frac{\partial}{\partial y}j+\frac{\partial}{\partial z}k $$ $$ \nabla f=\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j+\frac{\partial f}{\partial z}k $$ So we can therefore derive that the directional derivative $(\frac{\partial f}{\partial \mathbf{u}})$ is: $$ \frac{\partial f}{\partial \mathbf{u}}=\nabla f\cdot \mathbf{u} $$ Which equals: $$ \frac{\partial f}{\partial \mathbf{u}}(M)=|\nabla f|\cos\phi $$
The Lagrange multiplier technique lets you find the maximum or minimum of a multivariable function when there is some constraint on the input values you are allowed to use. $$ \phi(M)=f(M)+\lambda_1F_1(M)+...+\lambda_kF_k(M) $$ Note: See Phys113-Series and Differential Equations, towards the end for more on this.