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:
\[ \cot x = \frac{\cos x}{\sin x} \]and the quotient rule, we have:
\[ \frac{d}{dx} \cot x = \frac{\left(\frac{d}{dx} \cos x\right) \cdot \sin x - \cos x \cdot \left(\frac{d}{dx} \sin x\right)}{\sin^2 x} \]Substitute \( \frac{d}{dx} \cos x = -\sin x \) and \( \frac{d}{dx} \sin x = \cos x \):
\[ \frac{d}{dx} \cot x = \frac{(-\sin x)(\sin x) - \cos x (\cos x)}{\sin^2 x} = -\frac{\sin^2 x + \cos^2 x}{\sin^2 x} = -\csc^2 x \]
Formula:
\[ \frac{d}{dx} \cot x = -\csc^2 x \]
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.
-and-its-derivative.gif)
Derivative of the Composite Function cot(u(x))
Using the chain rule:
\[ \frac{d}{dx} \cot(u(x)) = \frac{d}{du}(\cot u) \cdot \frac{du}{dx} = -\csc^2(u) \cdot \frac{du}{dx} \]
Chain Rule Formula:
\[ \frac{d}{dx} \cot(u(x)) = -\csc^2(u(x))\,u'(x) \]
Examples with Solutions
Worked Examples: Differentiating Composite Cotangent Functions
Find the derivatives:
- \( f(x) = \cot(x^3 - 2x + 2) \)
- \( g(x) = \cot(e^x) \)
- \( h(x) = \cot\left(\frac{-2}{x^3 + 2}\right) \)
Solutions:
-
Let \( u = x^3 - 2x + 2 \), then \( u' = 3x^2 - 2 \).
\[ f'(x) = -\csc^2(u) \cdot u' = -(3x^2 - 2)\csc^2(x^3 - 2x + 2) \] -
Let \( u = e^x \), then \( u' = e^x \).
\[ g'(x) = -\csc^2(e^x) \cdot e^x = -e^x \csc^2(e^x) \] -
Let \( u = \frac{-2}{x^3+2} \). Using the quotient rule or power rule:
\[ u' = \frac{0 - (-2)(3x^2)}{(x^3+2)^2} = \frac{6x^2}{(x^3+2)^2} \] \[ h'(x) = -\csc^2\left(\frac{-2}{x^3+2}\right) \cdot \frac{6x^2}{(x^3+2)^2} = -\frac{6x^2}{(x^3+2)^2} \csc^2\left(\frac{2}{x^3+2}\right) \]