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 xA 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
Solution to Example 1
More References and linksRules of Differentiation of Functions in Calculus.Trigonometric Identities and Formulas. Derivatives of the Trigonometric Functions. Chain Rule of Differentiation in Calculus. |