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.
-and-its-derivative.gif)
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
- \( f(x) = \csc (-x^3+3) \)
- \( g(x) = \csc (\cos(x)) \)
- \( h(x) = \csc (\dfrac{1}{x^2+1}) \)
Solution to Example 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
\( \displaystyle \dfrac{d}{dx} f(x) = - \cot u \csc u \dfrac{d}{dx} u = - \cot (-x^3+3) \csc (-x^3+3) \times (-3x^2) \)
\( = 3x^2 \; \cot (-x^3+3) \; \csc (-x^3+3) \)
-
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
\( \displaystyle \dfrac{d}{dx} g(x) = - \cot u \csc u \dfrac{d}{dx} u = - \cot (\cos x) \csc (\cos x) \times (- \sin x) \)
\( = \sin x \;\cot (\cos x) \; \csc (\cos x) \)
-
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
\( \displaystyle \dfrac{d}{dx} h(x) = - \cot u \csc u \dfrac{d}{dx} u = - \cot (\dfrac{1}{x^2+1}) \csc (\dfrac{1}{x^2+1}) \times (-\dfrac{2x}{(x^2+1)^2}) \)
\( = \dfrac{2x}{(x^2+1)^2} \; \cot (\dfrac{1}{x^2+1}) \;\csc (\dfrac{1}{x^2+1}) \)
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 .
