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Derivative of cot(x)

Derivative of cot(x) using First Principle of Derivatives

Using the First Principle of Derivatives, we will prove that the derivative of cot⁡(x)\cot(x) is equal to −1/sin⁡2(x)-1/\sin^2(x). The derivative of cotangent is easier to prove if we take its identity, which is the inverse of the tangent.

Proof. Let f(x)=cot⁡(x)=1tan⁡(x)=cos⁡(x)sin⁡(x)f(x) = \cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}. Then
f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0cot⁡(x+h)−cot⁡(x)h=lim⁡h→0cos⁡(x+h)sin⁡(x+h)−cos⁡(x)sin⁡(x)h=lim⁡h→0cos⁡(x+h)sin⁡(x)−sin⁡(x+h)cos⁡(x)sin⁡(x+h)sin⁡(x)h\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*}
We will take the following identity
sin⁡(A−B)=sin⁡(A)cos⁡(B)−cos⁡(A)sin⁡(B).\begin{align*} \sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B). \end{align*}
Applying that to our case, then we get the following equality
cos⁡(x+h)sin⁡(x)−sin⁡(x+h)cos⁡(x)=sin⁡(−h).\begin{align*} \cos(x + h)\sin(x) - \sin(x + h)\cos(x) = \sin(-h). \end{align*}
Since lim⁡h→0sin⁡(−h)h=−lim⁡h→0sin⁡(h)h=−1\lim_{h \rightarrow 0} \frac{\sin(-h)}{h} = -\lim_{h \rightarrow 0} \frac{\sin(h)}{h} = -1 which we have seen here, we get:
lim⁡h→0sin⁡(−h)sin⁡(x+h)sin⁡(x)h=lim⁡h→0sin⁡(−h)h⋅lim⁡h→01sin⁡(x+h)sin⁡(x)=−lim⁡h→01sin⁡(x+h)sin⁡(x)=−1sin⁡2(x)=−csc⁡2(x).\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*}
Therefore, we have that f′(x)=−1sin⁡2(x)=−csc⁡2(x)f'(x) = - \frac{1}{\sin^2(x)} = -\csc^2(x).
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