# Show d/dx (cot x)=-(cosec x)^2.?

Mar 5, 2018

See explanation

#### Explanation:

We want find the derivative of

$y = \cot \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$

Use the quotient rule, if $y = \frac{f}{g}$

then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \cdot g - f \cdot g '}{g} ^ 2$, thus

• $f = \cos \left(x\right) \implies f ' = - \sin \left(x\right)$
• $g = \sin \left(x\right) \implies g ' = \cos \left(x\right)$

Thus

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \sin \left(x\right) \cdot \sin \left(x\right) - \cos \left(x\right) \cos \left(x\right)}{\sin} ^ 2 \left(x\right)$

$= \frac{- \left({\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)\right)}{\sin} ^ 2 \left(x\right)$

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

Or

$\frac{d}{\mathrm{dx}} \cot \left(x\right) = - {\csc}^{2} \left(x\right)$

Mar 5, 2018

$\text{see explanation}$

#### Explanation:

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{given "f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \textcolor{b l u e}{\text{quotient rule}}$

$f \left(x\right) = \cot x = \cos \frac{x}{\sin} x$

$g \left(x\right) = \cos x \Rightarrow g ' \left(x\right) = - \sin x$

$h \left(x\right) = \sin x \Rightarrow h ' \left(x\right) = \cos x$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\cot x\right)$

$= \frac{- {\sin}^{2} x - {\cos}^{2} x}{{\sin}^{2} x}$

$= \frac{- \left({\sin}^{2} x + {\cos}^{2} x\right)}{\sin} ^ 2 x$

$= - \frac{1}{\sin} ^ 2 x = - {\csc}^{2} x$