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

$f ' \left(x\right) = \frac{{x}^{4} \cdot \sin x + 2 {x}^{3} \cdot \cos x - 6 {x}^{2}}{x \cdot \cos x - 2} ^ 2$

#### Explanation:

the quotient rule formula $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{\mathrm{du}}{\mathrm{dx}} - u \cdot \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

the given: differentiate $f \left(x\right) = {x}^{3} / \left(x \cdot \cos x - 2\right)$

$f ' \left(x\right) = \frac{\left(x \cdot \cos x - 2\right) \left(3 {x}^{2}\right) - {x}^{3} \cdot \left(x \cdot \left(- \sin x\right) + \left(\cos x\right) \cdot \left(1\right) - 0\right)}{x \cdot \cos x - 2} ^ 2$

$f ' \left(x\right) = \frac{3 {x}^{3} \cdot \cos x - 6 {x}^{2} + {x}^{4} \cdot \sin x - {x}^{3} \cdot \cos x + 0}{x \cdot \cos x - 2} ^ 2$

$f ' \left(x\right) = \frac{\left(3 {x}^{3} - {x}^{3}\right) \cdot \cos x + {x}^{4} \cdot \sin x - 6 {x}^{2}}{x \cdot \cos x - 2} ^ 2$

$f ' \left(x\right) = \frac{2 {x}^{3} \cdot \cos x + {x}^{4} \cdot \sin x - 6 {x}^{2}}{x \cdot \cos x - 2} ^ 2$

the final simplification

$f ' \left(x\right) = \frac{{x}^{4} \cdot \sin x + 2 {x}^{3} \cdot \cos x - 6 {x}^{2}}{x \cdot \cos x - 2} ^ 2$