# Differentiate the following function? y = (8 x)/ (6 - cot x)

Feb 28, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 \frac{\left(6 - \cot x - x {\csc}^{2} x\right)}{6 - \cot x} ^ 2$

#### Explanation:

We apply the Quotient Rule for Differentiation:

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\text{ } \left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with $y = \frac{8 x}{6 - \cot x}$ Then

$\left\{\begin{matrix}\text{Let "u=8x & => & (du)/dx=8 \\ "And } v = 6 - \cot x & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = {\csc}^{2} x\end{matrix}\right.$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$\text{ } = \frac{\left(6 - \cot x\right) \left(8\right) - \left(8 x\right) \left({\csc}^{2} x\right)}{6 - \cot x} ^ 2$
$\text{ } = 8 \frac{\left(6 - \cot x - x {\csc}^{2} x\right)}{6 - \cot x} ^ 2$