Proof of the Derivative of cot(x)

The derivative of \( \cot(x) \) can be computed using the derivatives of \( \sin x \) and \( \cos x \) with the quotient rule. Derivatives of composite cotangent functions are also presented with worked examples.

Proof of the Derivative of cot(x)

Using the trigonometric identity:

and the quotient rule, we have:

Substitute \( \frac{d}{dx} \cos x = -\sin x \) and \( \frac{d}{dx} \sin x = \cos x \):

\[ \boxed{\frac{d}{dx} \cot x = -\csc^2 x} \]

Graph of cot(x) and Its Derivative

The graphs of \( \cot x \) and its derivative are shown below. Since \( \cot x \) is decreasing, its derivative is negative everywhere.

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

Using the chain rule:

\[ \boxed{\frac{d}{dx} \cot(u(x)) = -\csc^2(u(x))\,u'(x)} \]

Examples

Find the derivatives:

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

Solutions

1. \[ u = x^3 - 2x + 2, \quad u' = 3x^2 - 2 \] \[ f'(x) = -\csc^2(u) \cdot u' = -(3x^2-2)\csc^2(x^3-2x+2) \]
2. \[ u = e^x, \quad u' = e^x \] \[ g'(x) = -\csc^2(e^x) \cdot e^x = -e^x \csc^2(e^x) \]
3. \[ u = \frac{-2}{x^3+2}, \quad u' = \frac{6x^2}{(x^3+2)^2} \] \[ h'(x) = -\csc^2\left(\frac{-2}{x^3+2}\right) \cdot \frac{6x^2}{(x^3+2)^2} = -\frac{6x^2}{(x^3+2)^2} \csc^2\left(\frac{2}{x^3+2}\right) \]

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