# How do you differentiate f(x)=x / cot (x) + 3 using the quotient rule?

May 28, 2016

$\setminus \frac{x + \setminus {\sin}^{2} t \left(x\right) \setminus \cot \left(x\right)}{\setminus {\sin}^{2} \setminus \left(x\right) \setminus {\cot}^{2} \setminus \left(x\right)}$

#### Explanation:

$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \frac{x}{\setminus \cot \left(x\right)} + 3 \setminus\right)$

Applying sum/difference rule,$\setminus {\left(f \setminus \pm g \setminus\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$

$= \setminus \frac{d}{\mathrm{dx}} \setminus \left(\setminus \frac{x}{\setminus \cot t \left(x\right)}\right) + \setminus \frac{d}{\mathrm{dx}} \left(3\right)$

Now,
$= \setminus \frac{d}{\mathrm{dx}} \setminus \left(\setminus \frac{x}{\setminus \cot t \left(x\right)}\right)$
Applying quotient rule,${\left(\setminus \frac{f}{g}\right)}^{'} = \setminus \frac{{f}^{'} \setminus \cdot g - {g}^{'} \setminus \cdot f}{{g}^{2}}$

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

And,we know,$\setminus \frac{d}{\mathrm{dx}} \left(x\right) = 1$ and, $\setminus \frac{d}{\mathrm{dx}} \left(\setminus \cot \left(x\right)\right) = - \setminus \frac{1}{\setminus {\sin}^{2} \left(x\right)}$

Also,$\setminus \frac{d}{\mathrm{dx}} \left(3\right) = 0$

Finally,
$= \setminus \frac{\setminus {\sin}^{2} \left(x\right) \setminus \cot \left(x\right) + x}{\setminus {\sin}^{2} \setminus \left(x\right) \setminus {\cot}^{2} \left(x\right)} + 0$

Simplifying it,
$\setminus \frac{x + \setminus {\sin}^{2} t \left(x\right) \setminus \cot \left(x\right)}{\setminus {\sin}^{2} \setminus \left(x\right) \setminus {\cot}^{2} \setminus \left(x\right)}$