 # Proof of Derivative of cot x

The derivative of $\cot (x)$ is computed using the derivative of $\sin x$ and $\cos x$ and the quotient rule of differentiation. Examples of derivatives of cotangent composite functions are also presented along with their solutions.

## Proof of the Derivative of cot x

A trigonometric identity relating $\cot x$, $\cos x$ and $\sin x$ is given by $\cot x = \dfrac { \cos x }{ \sin x }$ We now use the quotient rule of differentiation way to find the derivative of $\cot x$
$\displaystyle {\dfrac {d}{dx} \cot x = \dfrac {d}{dx} (\dfrac{\cos x }{\sin x}) = \dfrac{{ (\dfrac {d}{dx}\cos x) }{ \sin x } - \cos x (\dfrac {d}{dx} \sin x) }{\sin^2 x}}$

Use the formulae for the
derivative of the trigonometric functions $\sin x$ and $\cos x$ given by $\dfrac {d}{dx}\cos x = - \sin x$ and $\dfrac {d}{dx}\sin x = \cos x$ and substitute to obtain

$\displaystyle {\dfrac {d}{dx} \cot x = \dfrac{{ -\sin x \sin x } - \cos x \cos x }{\sin^2 x}}$

Simplify

$\displaystyle {= - \dfrac{ \sin^2 x + \cos^2 x } {\sin^2 x} = - \dfrac{ 1 }{\sin^2 x} = - \csc^2 x }$
conclusion
$\displaystyle {\dfrac {d}{dx} \cot x = - \csc^2 x}$

## Graph of cot x and its Derivative

The graphs of $\cot(x)$ and its derivative are shown below. The derivative of cot(x) is negative everywhere because cot(x) is a decreasing function. ## Derivative of the Composite Function cot (u(x))

We now consider a composite function which is a function cot of another function u. Use the chain rule of differentiation to write

$\displaystyle \dfrac{d}{dx} \cot (u(x)) = (\dfrac{d}{du} \cot u) (\dfrac{d}{dx} u )$

Simplify

$= - \csc^2 u \dfrac{d}{dx} u$

Conclusion

$\displaystyle \dfrac{d}{dx} \cot (u(x)) = - \csc^2 u \dfrac{d}{dx} u$

Example
Find the derivative of the composite tan functions

1. $f(x) = \cot (x^3-2x+2)$
2. $g(x) = \cot (e^x)$
3. $h(x) = \cot (\dfrac{-2}{x^3+2})$

Solution to the Above Example

1. Let $u(x) = x^3-2x+2$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} (x^3-2x+2) = 3x^2-2$ and apply the rule for the composite cot function given above.

$\displaystyle \dfrac{d}{dx} f(x) = -\csc^2 u \dfrac{d}{dx} u = -\csc^2 (x^3-2x+2) \times (3x^2-2)$
$= -(3x^2-2) \csc^2 (x^3-2x+2)$

2. Let $u(x) = e^x$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} e^x = e^x$ and apply the rule of cot composite given above.

$\displaystyle \dfrac{d}{dx} g(x) = - \csc^2 u \dfrac{d}{dx} u$
$= - \csc^2 (e^x) \times e^x = - e^x \csc^2 (e^x)$

3. Let $u(x) = \dfrac{-2}{x^3+2}$ and therefore $\dfrac{d}{dx} u = \dfrac{6x^2}{\left(x^3+2\right)^2}$ and apply the rule of cot composite obtained above

$\displaystyle \dfrac{d}{dx} h(x) = - \csc^2 u \dfrac{d}{dx} u = - \csc^2 (\dfrac{-2}{x^3+2}) \times \dfrac{6x^2}{\left(x^3+2\right)^2}$
$= - \dfrac{6x^2}{\left(x^3+2\right)^2} \csc^2 (\dfrac{-2}{x^3+2}) = - \dfrac{6x^2}{\left(x^3+2\right)^2} \csc^2 (\dfrac{2}{x^3+2})$ 