The derivative of \( \csc(x) \) can be computed using the quotient rule. Derivatives of composite cosecant functions are also presented with worked examples.
Proof of the Derivative of csc(x)
Using the trigonometric identity:
\[ \csc x = \frac{1}{\sin x} \]and the quotient rule:
\[ \frac{d}{dx}\csc x = \frac{\left(\frac{d}{dx} 1\right) \cdot \sin x - 1 \cdot \left(\frac{d}{dx} \sin x\right)}{\sin^2 x} \]Since \( \frac{d}{dx} 1 = 0 \) and \( \frac{d}{dx} \sin x = \cos x \), we get:
\[ \frac{d}{dx}\csc x = \frac{0 - \cos x}{\sin^2 x} = -\frac{\cos x}{\sin^2 x} = -\cot x \, \csc x \]
Formula:
\[ \frac{d}{dx}\csc x = -\cot x \, \csc x \]
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))
Using the chain rule:
\[ \frac{d}{dx} \csc(u(x)) = \frac{d}{du}(\csc u) \cdot \frac{du}{dx} = -\cot u \, \csc u \cdot \frac{du}{dx} \]
Chain Rule Formula:
\[ \frac{d}{dx} \csc(u(x)) = -\cot(u(x)) \, \csc(u(x)) \, u'(x) \]
Examples with Solutions
Worked Examples: Differentiating Composite Cosecant Functions
Find the derivatives:
- \( f(x) = \csc(-x^3 + 3) \)
- \( g(x) = \csc(\cos x) \)
- \( h(x) = \csc\left(\frac{1}{x^2+1}\right) \)
Solutions:
-
Let \( u = -x^3 + 3 \), then \( u' = -3x^2 \).
\[ f'(x) = -\cot(u) \, \csc(u) \, u' = -\cot(-x^3+3) \, \csc(-x^3+3) \cdot (-3x^2) = 3x^2 \, \cot(-x^3+3) \, \csc(-x^3+3) \] -
Let \( u = \cos x \), then \( u' = -\sin x \).
\[ g'(x) = -\cot(u) \, \csc(u) \, u' = -\cot(\cos x) \, \csc(\cos x) \cdot (-\sin x) = \sin x \, \cot(\cos x) \, \csc(\cos x) \] -
Let \( u = \frac{1}{x^2+1} \), then \( u' = -\frac{2x}{(x^2+1)^2} \).
\[ h'(x) = -\cot(u) \, \csc(u) \, u' = -\cot\left(\frac{1}{x^2+1}\right) \, \csc\left(\frac{1}{x^2+1}\right) \cdot \left(-\frac{2x}{(x^2+1)^2}\right) = \frac{2x}{(x^2+1)^2} \, \cot\left(\frac{1}{x^2+1}\right) \, \csc\left(\frac{1}{x^2+1}\right) \]