# How do you differentiate f(x)=(-2x^2-5)/cos(2x^3)?

Jan 15, 2017

$f ' \left(x\right) = \frac{- 4 x \cos \left(2 {x}^{3}\right) - \left(12 {x}^{4} + 5\right) \sin \left(2 {x}^{3}\right)}{\cos} ^ 2 \left(2 {x}^{3}\right)$

#### Explanation:

Using the chain rule for the denominator:

$\left(A \left(B \left(x\right)\right)\right) ' = A ' \left(B \left(x\right)\right) \cdot B ' \left(x\right)$

$\left(\cos \left(2 {x}^{3}\right)\right) ' = - \sin \left(2 {x}^{3}\right) \cdot 6 {x}^{2}$

Then, using the quotient rule:

$\left(\frac{A \left(x\right)}{B \left(x\right)}\right) ' = \frac{A ' \left(x\right) B \left(x\right) - A \left(x\right) B ' \left(x\right)}{{B}^{2} \left(x\right)}$

$\frac{- 4 x \cos \left(2 {x}^{3}\right) - \left(- \sin \left(2 {x}^{3}\right) \cdot 6 {x}^{2}\right) \left(- 2 {x}^{2} - 5\right)}{\cos} ^ 2 \left(2 {x}^{3}\right)$

=(-4xcos(2x^3) - (12x^4sin(2x^3) + 5sin(2x^3)))/(cos^2(2x^3)

$= \frac{- 4 x \cos \left(2 {x}^{3}\right) - \left(12 {x}^{4} + 5\right) \sin \left(2 {x}^{3}\right)}{\cos} ^ 2 \left(2 {x}^{3}\right)$