**Proof.**Let f(x) = \cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}. Then

\begin{align*} f'(x) &= \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \rightarrow 0} \frac{\cot(x + h) - \cot(x)}{h} \\ &= \lim_{h \rightarrow 0} \frac{\frac{\cos(x + h)}{\sin(x + h)} - \frac{\cos(x)}{\sin(x)}}{h} \\ &= \lim_{h \rightarrow 0} \frac{\frac{\cos(x + h)\sin(x) - \sin(x + h)\cos(x)}{\sin(x + h)\sin(x)}}{h} \end{align*}

\begin{align*} \sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B). \end{align*}

\begin{align*} \cos(x + h)\sin(x) - \sin(x + h)\cos(x) = \sin(-h). \end{align*}

\begin{align*} \lim_{h \rightarrow 0} \frac{\frac{\sin(-h)}{\sin(x + h)\sin(x)}}{h} &= \lim_{h \rightarrow 0} \frac{\sin(-h)}{h} \cdot \lim_{h \rightarrow 0} \frac{1}{\sin(x + h)\sin(x)} \\ &= - \lim_{h \rightarrow 0} \frac{1}{\sin(x + h)\sin(x)} \\ &= - \frac{1}{\sin^2(x)} \\ &= - \csc^2(x). \end{align*}