Proof of the Derivative of csc x

The proof of the calculation of the derivative of $\csc (x)$ is presented using the quotient rule of derivatives.

Proof of the Derivative of csc x

A trigonometric identity relating $\csc x$ and $\sin x$ is given by $\csc x = \dfrac { 1 }{ \sin x }$ Use of the quotient rule of differentiation to find the derivative of $\csc x$; hence
$\displaystyle { \dfrac {d}{dx} \csc x = \dfrac {d}{dx} (\dfrac{ 1 }{\sin x}) = \dfrac { (\dfrac {d}{dx}1) { \sin x } - 1 (\dfrac {d}{dx} \sin x) } {\sin^2 x} }$

The derivative of 1 is equal to zero. Use the formulae for the
derivative of the trigonometric functions $\sin x$ given by $\dfrac {d}{dx}\sin x = \cos x$ and substitute to obtain

$\displaystyle {\dfrac {d}{dx} \csc x = \dfrac{ (0 - (\cos x) )}{\sin^2 x}}$

Simplify

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

conclusion
$\displaystyle {\dfrac {d}{dx} \csc x = - \cot x \; \csc x}$

Graph of csc x and its Derivative

The graphs of $\csc(x)$ and its derivative are shown below.

Derivative of the Composite Function csc (u(x))

Let us consider the composite function csc of another function u(x). Use the chain rule of differentiation to write

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

Simplify

$= - \cot u \csc u \dfrac{d}{dx} u$

Conclusion

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

Example 1
Find the derivative of the composite csc functions

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

Solution to Example 1

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

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

$= 3x^2 \; \cot (-x^3+3) \; \csc (-x^3+3)$

2. Let $u(x) = \cos x$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} \cos x = - \sin x$ and apply the above rule of differentiation for the composite csc function

$\displaystyle \dfrac{d}{dx} g(x) = - \cot u \csc u \dfrac{d}{dx} u = - \cot (\cos x) \csc (\cos x) \times (- \sin x)$

$= \sin x \;\cot (\cos x) \; \csc (\cos x)$

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

$\displaystyle \dfrac{d}{dx} h(x) = - \cot u \csc u \dfrac{d}{dx} u = - \cot (\dfrac{1}{x^2+1}) \csc (\dfrac{1}{x^2+1}) \times (-\dfrac{2x}{(x^2+1)^2})$

$= \dfrac{2x}{(x^2+1)^2} \; \cot (\dfrac{1}{x^2+1}) \;\csc (\dfrac{1}{x^2+1})$