# Calculus

## Derivative formulae

Trigonometry functionHyperbolic function
$\frac{d}{dx} \sin(x) = \cos(x)$$\frac{d}{dx} \sinh(x) = \cosh(x) \frac{d}{dx} \cos(x) = - \sin(x)$$\frac{d}{dx} \cosh(x) = \sinh(x)$
$\frac{d}{dx} \tan(x) = \sec^{2}(x)$$\frac{d}{dx} \tanh(x) = \text{sech}^{2}(x) \frac{d}{dx} \cot(x) = -\text{cosec}^{2}(x)$$\frac{d}{dx} \coth(x) = -\text{cosech}^{2}(x)$
$\frac{d}{dx} \text{cosec}(x) = -\text{cosec}(x)\cot(x)$$\frac{d}{dx} \text{cosech}(x) = -\text{cosech}(x)\coth(x) \frac{d}{dx} \sec(x) = \sec(x)\tan(x)$$\frac{d}{dx} \text{sech}(x) = -\text{sech}(x)\tanh(x)$
Inverse trigonometry functionInverse hyperbolic function
$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1 - x^{2}}}$$\frac{d}{dx} \sinh^{-1}(x) = \frac{1}{\sqrt{1 + x^{2}}} \frac{d}{dx} \cos^{-1}(x) = \frac{-1}{\sqrt{1 - x^{2}}}$$\frac{d}{dx} \cosh^{-1}(x) = \frac{1}{x^{2} - 1}$
$\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1 + x^{2}}$$\frac{d}{dx} \tanh^{-1}(x) = \frac{1}{1 - x^{2}} \frac{d}{dx} \cot^{-1}(x) = \frac{-1}{x^{2} + 1}$$\frac{d}{dx} \coth^{-1}(x) = \frac{-1}{x^{2} - 1}$
$\frac{d}{dx} \sec^{-1}(x) = \frac{1}{x \sqrt{x^{2} - 1}}$$\frac{d}{dx} \text{sech}^{-1}(x) = \frac{-1}{x \sqrt{1 - x^{2}}} \frac{d}{dx} \text{cosec}^{-1}(x) = \frac{-1}{x\sqrt{x^{2} - 1}}$$\frac{d}{dx} \text{cosech}^{-1}(x) = \frac{-1}{x \sqrt{x^{2} + 1}}$

Find the mnemonic for the derivative of inverse hyperbolic functions at here.

### Vector Derivative

Let $\mathbf{x}, \mathbf{u}$ be vectors of length $n$, and let $A$ be a matrix of size $n\times n$ then,

\begin{align} \frac{\partial}{\partial \mathbf{x}} (\mathbf{u}^{T}\mathbf{x}) & = \frac{\partial}{\partial \mathbf{x}} (\mathbf{x}^{T}\mathbf{u}) = \mathbf{u}^{T} \\ \frac{\partial}{\partial \mathbf{x}} (\mathbf{A}\mathbf{x}) & = \mathbf{A} \\ \frac{\partial}{\partial \mathbf{x}} (\mathbf{x}^{T} \mathbf{A} \mathbf{x}) & = \mathbf{x}^{T} (\mathbf{A} + \mathbf{A}^{T}) \\ \frac{\partial^{2}}{\partial \mathbf{x}^{2}} (\mathbf{x^{T}}\mathbf{A}\mathbf{x}) & = \mathbf{A} + \mathbf{A}^{T} \end{align}

Note that if $\mathbf{A}$ is a symmetric matrix then, $\mathbf{A} + \mathbf{A}^{T} = 2 \mathbf{A}$.

## Integration formulae

### Gamma function

• $\forall n \in \mathbb{Z}^{+}$, $\Gamma(n) = (n-1)!$
• Def. $\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt, \quad \Re(z)>0$.