Proof of the Derivative of csc(x)

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:

\[ \frac{d}{dx}\csc x = \frac{ (\frac{d}{dx} 1) \cdot \sin x - 1 \cdot (\frac{d}{dx} \sin x) }{\sin^2 x} \]

Since \( \frac{d}{dx} 1 = 0 \) and \( \frac{d}{dx} \sin x = \cos x \), we get

\[ \frac{d}{dx}\csc x = \frac{0 - \cos x}{\sin^2 x} = -\frac{\cos x}{\sin^2 x} = -\cot x \csc x \]

\[ \boxed{\frac{d}{dx}\csc x = -\cot x \, \csc x} \]

Graph of csc(x) and Its Derivative

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

Graph of csc x and its derivative

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

Using the chain rule:

\[ \frac{d}{dx} \csc(u(x)) = \frac{d}{du}(\csc u) \cdot \frac{du}{dx} = -\cot u \, \csc u \cdot \frac{du}{dx} \]

\[ \boxed{\frac{d}{dx} \csc(u(x)) = -\cot(u(x)) \, \csc(u(x)) \, u'(x)} \]

Examples

Find the derivatives:

  1. \( f(x) = \csc(-x^3 + 3) \)
  2. \( g(x) = \csc(\cos x) \)
  3. \( h(x) = \csc\left(\frac{1}{x^2+1}\right) \)

Solutions

  1. \[ u = -x^3 + 3, \quad u' = -3x^2 \] \[ f'(x) = -\cot(u) \, \csc(u) \, u' \\\\ = -\cot(-x^3+3) \, \csc(-x^3+3) \cdot (-3x^2) \\\\ = 3x^2 \, \cot(-x^3+3) \, \csc(-x^3+3) \]
  2. \[ u = \cos x, \quad u' = -\sin x \] \[ g'(x) = -\cot(u) \, \csc(u) \, u' \\\\= -\cot(\cos x) \, \csc(\cos x) \cdot (-\sin x) \\\\= \sin x \, \cot(\cos x) \, \csc(\cos x) \]
  3. \[ u = \frac{1}{x^2+1}, \quad u' = -\frac{2x}{(x^2+1)^2} \] \[ h'(x) = -\cot(u) \, \csc(u) \, u' \\\\ = -\cot\left(\frac{1}{x^2+1}\right) \, \csc\left(\frac{1}{x^2+1}\right) \cdot \left(-\frac{2x}{(x^2+1)^2}\right) \\\\ = \frac{2x}{(x^2+1)^2} \, \cot\left(\frac{1}{x^2+1}\right) \, \csc\left(\frac{1}{x^2+1}\right) \]

More References

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