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Calculating the derivative of cot(x)

Posted in Math Theory by Lia on 09/19/2017

Consider the function:

$\mathbf{f(x)=\cot x}$

Replacing the function f(x) in the general derivative formula, we get:

$\frac{\mathrm{d} f(x)}{\mathrm{d} x}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{(x+h)-x}=\lim_{h \to0 }\frac{\cot (x+h)-\cot x}{h}=$

$\lim_{h \to0 }\frac{ -\frac{\sin (x+h-x)}{\sin (x+h)\sin x}}{h}=\lim_{h \to0 }\frac{ -\frac{\sin (h)}{\sin (x+h)\sin x}}{h}=\lim_{h \to0 }{ \frac{-\sin (h)}{h*\sin (x+h)\sin x}}$

$-\lim_{h \to0 }{ \frac{\sin (h)}{h}}*\lim_{h \to0 }\frac{1}{\sin (x+h)*\sin x}=-\lim_{h \to0 }\frac{1}{\sin (x+h)*\sin x}=$

$-\lim_{h \to0 }\frac{1}{\frac{1}{2}(\cos (x+h-x)-cos(x+h+x))}=-\lim_{h \to0 }\frac{1}{\frac{1}{2}(\cos (h)-cos(2x+h))}=$

$-\frac{1}{\frac{1}{2}*(2\sin^{2} x)}=-\frac{1}{\sin ^{2}x}$

But:

$\lim_{h \to0 }\frac{\sin h}{h}=1$

Therefore:

$\frac{\mathrm{d}}{\mathrm{d} x}(\cot x)=-\frac{1}{\sin ^{2}x}$

Note:

$\textbf{When}\: \: x=k\frac{\pi }{2}\textbf{,} \: \: \sin x=0\: \: \textbf{therefore}\: \: \frac{\mathrm{d}}{\mathrm{d} x}(\cot x) \textbf{ is} \: \: \textbf{undefined.}$

$\textbf{Where} \: \textit{\textbf{k=constant}}.$

Tagged with: calculus, math theory

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