# How do you differentiate f(x)= ( 2 - xsecx )/ (x -3)  using the quotient rule?

Mar 23, 2017

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{\left(- x \sec \left(x\right) \tan \left(x\right) - \sec \left(x\right)\right) \left(x - 3\right) - \left(2 - x \sec x\right)}{x - 3} ^ 2$

#### Explanation:

(For this problem, I am using $h \left(x\right)$ where $f \left(x\right)$ is traditionally used, to avoid ambiguity since the original function is named $f \left(x\right)$)

The quotient rule states that:

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} \frac{h \left(x\right)}{g} \left(x\right) = \frac{h ' \left(x\right) g \left(x\right) - h \left(x\right) g ' \left(x\right)}{g} {\left(x\right)}^{2}$

In this case, we can say that:

$h \left(x\right) = 2 - x \sec x$
$g \left(x\right) = x - 3$

Therefore, using standard differentiation rules:

$h ' \left(x\right) = - x \sec \left(x\right) \tan \left(x\right) - \sec \left(x\right)$
$g ' \left(x\right) = 1$

Now we can plug these values into the quotient rule formula:

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{\left(- x \sec \left(x\right) \tan \left(x\right) - \sec \left(x\right)\right) \left(x - 3\right) - \left(2 - x \sec x\right)}{x - 3} ^ 2$

This is the final form of the derivative, but it can be expanded and simplified if necessary.