# What is the derivative of y=cot(x) using the first principal rule?

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#### Explanation

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#### Explanation:

I want someone to double check my answer

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Jim H Share
Nov 20, 2016

See the explanation.

#### Explanation:

Let $f \left(x\right) = \cot x$, then , by definition, $f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\cot \left(x + h\right) - \cot x}{h}$.

Well try to simplify using $\cot \theta = \cos \frac{\theta}{\sin} \theta$ and also using the sum laws for sine and for cosine.

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\cot \left(x + h\right) - \cot x}{h}$

$= {\lim}_{h \rightarrow 0} \frac{\cos \frac{x + h}{\sin} \left(x + h\right) - \cos \frac{x}{\sin} x}{h}$

 = lim_(hrarr0)(sinxcos(x+h) - cosxsin(x+h))/(hsin(x+h)sinx

 = lim_(hrarr0)(sinx[cosxcos h-sinxsin h] - cosx[sinxcos h+cosxsin h])/(hsin(x+h)sinx

 = lim_(hrarr0)(-sin^2xsin h - cos^2xsin h)/(hsin(x+h)sinx

 = lim_(hrarr0)(-sin h)/(hsin(x+h)sinx

$= {\lim}_{h \rightarrow 0} - \sin \frac{h}{h} \frac{1}{\sin \left(x + h\right) \sin x}$

$= - \left(1\right) \frac{1}{\sin x \sin x} = - \frac{1}{\sin} ^ 2 x = - {\csc}^{2} x$

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