# How do you differentiate f(x)= ( 4- 3 secx )/ (3x + 1)  using the quotient rule?

Sep 7, 2016

$f ' \left(x\right) = \frac{- 9 x \tan x \sec x - 3 \tan x \sec x - 12 + 9 \sec x}{9 {x}^{2} + 6 x + 1}$

#### Explanation:

Let's start by finding the derivative of the numerator, since it is somewhat complex and we will need it to apply the quotient rule.

$\frac{d}{\mathrm{dx}} \left(4 - 3 \sec x\right) = \frac{d}{\mathrm{dx}} \left(4\right) - \frac{d}{\mathrm{dx}} \left(\frac{3}{\cos} x\right) = 0 - \frac{0 \times \cos x - \left(- \sin x \times 3\right)}{\cos} ^ 2 x = - \frac{3 \sin x}{\cos} ^ 2 x = - 3 \tan x \sec x$

We can now go on with the quotient rule on $f \left(x\right)$.

$f ' \left(x\right) = \frac{- 3 \tan x \sec x \left(3 x + 1\right) - 3 \left(4 - 3 \sec x\right)}{3 x + 1} ^ 2$

$f ' \left(x\right) = \frac{- 9 x \tan x \sec x - 3 \tan x \sec x - 12 + 9 \sec x}{9 {x}^{2} + 6 x + 1}$

Hopefully this helps!