# How do you differentiate (-2x^2 -5x-4 )/ (cos2x) using the quotient rule?

Jun 8, 2017

$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{\cos 2 x \left(4 x + 5\right) - 2 \sin 2 x \left(2 {x}^{2} + 5 x + 4\right)}{{\cos}^{2} 2 x}$

#### Explanation:

Quotient rule states if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right) - \frac{\mathrm{dh}}{\mathrm{dx}} \times g \left(x\right)}{h \left(x\right)} ^ 2$

Hence, as $f \left(x\right) = \frac{- 2 {x}^{2} - 5 x - 4}{\cos 2 x}$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\cos 2 x \left(- 4 x - 5\right) - \left(- 2 \sin 2 x\right) \left(- 2 {x}^{2} - 5 x - 4\right)}{{\cos}^{2} 2 x}$

= $\frac{- 4 x \cos 2 x - 5 \cos 2 x - 2 \sin 2 x \left(2 {x}^{2} + 5 x + 4\right)}{{\cos}^{2} 2 x}$

= $\frac{- 4 x \cos 2 x - 5 \cos 2 x - 4 {x}^{2} \sin 2 x - 10 x \sin 2 x - 8 \sin 2 x}{{\cos}^{2} 2 x}$

= $- \frac{\cos 2 x \left(4 x + 5\right) - 2 \sin 2 x \left(2 {x}^{2} + 5 x + 4\right)}{{\cos}^{2} 2 x}$