Calculus cheat sheet

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}}$

Mnemonic to remember derivative of inverse hyperbolic function

  • First of all, remember the derivative of inverse trigonometry function exactly how I written in RHS.
  • In order to memorize the formulae for the derivative of inverse hyperbolic function, use the below rule:
    • For $\cosh^{-1}$ and $\text{sech}^{-1}$, multiply by minus in the numerator (up) and inside the square root in the denominator (down) of $\cosh^{-1}$ and $\text{sech}^{-1}$ respectively i.e. $\frac{\text{up}}{\sqrt{\text{down}}} \to \frac{\text{- up}}{\sqrt{\text{- down}}}$.
    • For others, change the sign of second term in the denominator. i.e. $\pm \to \mp$

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, $\mathbf{A} + \mathbf{A}^{T} = 2 \mathbf{A}$.

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Published on May 7, 2021

Last revised on May 9, 2021