# How do you differentiate f(x)= ( x + 1 )/ ( tan x ) using the quotient rule?

Mar 6, 2016

$f ' \left(x\right) = \cot x - x {\csc}^{2} x - {\csc}^{2} x$

#### Explanation:

The quotient rule states that:
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$
Where $u$ and $v$ are functions of $x$.

We use this whenever we need to find the derivative of one function divided by another, which is the case here. We can say let $u = x + 1$ and $v = \tan x$; finding the derivatives of each of these individually:
$u ' = 1$
$v ' = {\sec}^{2} x$

Substituting these into the quotient rule formula, we have:
$f ' \left(x\right) = \frac{\left(x + 1\right) ' \left(\tan x\right) - \left(x + 1\right) \left(\tan x\right) '}{\tan x} ^ 2$
$f ' \left(x\right) = \frac{\tan x - \left(x + 1\right) \left({\sec}^{2} x\right)}{{\tan}^{2} x}$

Now we move onto the algebra/trig, which tends to be the most difficult part:
$f ' \left(x\right) = \tan \frac{x}{\tan} ^ 2 x - \frac{\left(x + 1\right) \left({\sec}^{2} x\right)}{{\tan}^{2} x}$
$f ' \left(x\right) = \cot x - \frac{\left(x + 1\right) \left(\frac{1}{\cos} ^ 2 x\right)}{{\sin}^{2} \frac{x}{\cos} ^ 2 x}$
$f ' \left(x\right) = \cot x - \left(x + 1\right) \left(\frac{1}{\cos} ^ 2 x\right) \cdot \left({\cos}^{2} \frac{x}{\sin} ^ 2 x\right)$
$f ' \left(x\right) = \cot x - \left(x + 1\right) \left({\csc}^{2} x\right)$
$f ' \left(x\right) = \cot x - \left(x {\csc}^{2} x + {\csc}^{2} x\right)$
$f ' \left(x\right) = \cot x - x {\csc}^{2} x - {\csc}^{2} x$

We could do more simplifying here, but this result suffices.