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Expert-verified Found in: Page 232 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 Section Derivatives of Trigonometric and Hyperbolic Functions states the derivative of Trigonometric Functions, Inverse Trigonometric Functions, Hyperbolic Functions, and Inverse Hyperbolic Functions.

See the step by step solution

## Step 1. Given information.

The given section topic is Derivatives of Trigonometric and Hyperbolic Functions.

## Step 2. Derivatives of the Trigonometric Functions

Derivatives of the Trigonometric Functions are follows

$\frac{d}{dx}\left(\mathrm{sin}x\right)=\mathrm{cos}x\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\mathrm{cos}x\right)=-\mathrm{sin}x\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\mathrm{tan}x\right)=se{c}^{2}x\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(secx\right)=secx\mathrm{tan}x\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(cscx\right)=-cscxcotx\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(cotx\right)=-cs{c}^{2}x$

## Step 3. Derivatives of Inverse Trigonometric Functions

Derivatives of Inverse Trigonometric Functions are following.

$\frac{d}{dx}{\mathrm{sin}}^{-1}x=\frac{1}{\sqrt{1-{x}^{2}}}\phantom{\rule{0ex}{0ex}}\frac{d}{dx}{\mathrm{tan}}^{-1}x=\frac{1}{1+{x}^{2}}\phantom{\rule{0ex}{0ex}}\frac{d}{dx}se{c}^{-1}x=\frac{1}{\left|x\right|\sqrt{{x}^{2}-1}}$

## Step 4. Derivatives of Hyperbolic Functions and Inverse Hyperbolic Functions.

Derivatives of Hyperbolic Functions are following.

$\frac{d}{dx}\left(\mathrm{sin}hx\right)=\mathrm{cos}hx\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\mathrm{cos}hx\right)=\mathrm{sin}hx\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\mathrm{tan}hx\right)=sec{h}^{2}x$

Derivatives of Inverse Hyperbolic Functions are following.
$\frac{d}{dx}\mathrm{sin}{h}^{-1}x=\frac{1}{\sqrt{{x}^{2}+1}}\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\mathrm{cos}{h}^{-1}x=\frac{1}{\sqrt{{x}^{2}-1}}\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\mathrm{tan}{h}^{-1}x=\frac{1}{1-{x}^{2}}$ ### Want to see more solutions like these? 