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.

Graph of cot x and its derivative

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

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


More References and links

Rules of Differentiation of Functions in Calculus .
Trigonometric Identities and Formulas .
Derivatives of the Trigonometric Functions .
Chain Rule of Differentiation in Calculus .