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

Feb 11, 2016

$f ' \left(x\right) = \frac{3 \left(\left(3 - x\right) \sec x \tan x + \sec x - 1\right)}{x - 3} ^ 2$

#### Explanation:

The quotient rule states that

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

The respective derivatives contained within are $\frac{d}{\mathrm{dx}} \left(x - 3 \sec x\right) = 1 - 3 \sec x \tan x$ and $\frac{d}{\mathrm{dx}} \left(x - 3\right) = 1$. We obtain

$f ' \left(x\right) = \frac{\left(x - 3\right) \left(1 - 3 \sec x \tan x\right) - \left(x - 3 \sec x\right) \left(1\right)}{x - 3} ^ 2$

We can distribute and simplify.

$f ' \left(x\right) = \frac{x - 3 - 3 x \sec x \tan x + 9 \sec x \tan x - x + 3 \sec x}{x - 3} ^ 2$

$f ' \left(x\right) = \frac{3 \left(\left(3 - x\right) \sec x \tan x + \sec x - 1\right)}{x - 3} ^ 2$