# Lars Eighner's Homepage

## LarsWiki

### The Derivative of Arc Cotangent

Cotangent

##### Contents

A previous result is: {$${d \over {d\theta}} \cot\theta = - \csc^2\theta$$}

Attention is called to the diagram which illustrates the Pythagorean identity:

{$$csc^2\theta = 1 + cot^2\theta$$}

{$y = \cot(x)$} and {$y=\operatorname{arccot}(x)$}

The demonstration proceeds by implicit differentiation:

{\begin{align} \cot(\operatorname{arccot}\theta) &= \theta \tag{definition of inverse} \cr \text{Let } y &= \operatorname{arccot}\theta \cr \cot(y) &= \theta \cr {d \over {d\theta}} \left( \cot(y) \right) &= {d \over {d\theta}} \theta \cr {d \over {d\theta}}cot(y){d \over {d\theta}}y &= 1 \tag{Chain rule} \cr -\csc^2{d \over {d\theta}}y &= 1 \tag{previous result} \cr {d \over {d\theta}}y &= - {1 \over {csc^2(y)}} \cr {d \over {d\theta}}y & = - {1 \over {1 + cot^2(y)}} \tag{Pythagorean identity} \cr {d \over {d\theta}}\operatorname{arccot}\theta &= - {1 \over {1 + cot^2(\operatorname{arccot}\theta)}} \tag{value y} \cr \therefore \quad {d \over {d\theta}}\operatorname{arccot}\theta &= - {1 \over {1 + \theta^2}}\end{align}}

{$y = \cot(x)$}, {$y=\operatorname{arccot}(x)$}, {$y = \cot(x)$} and {$y=\operatorname{arccot}(x)$}.

Sources:

Recommended:

Category: Math Calculus Trigonometry

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

Figures are often enhanced by hand editing; the same results may not be achieved with source sites and source apps.

### August 05, 2017

• HomePage
• WikiSandbox

Lars

Contact by Snail!

Lars Eighner
APT 1191
8800 N IH 35
AUSTIN TX 78753
USA

Help