# What are the first three derivatives of (xcos(x)-sin(x))/(x^2)?

Mar 12, 2015

$y ' ' = \frac{- {x}^{3} \cos x + 3 {x}^{4} \sin x + 6 x \cos x - 6 \sin x}{x} ^ 4$.

This is why:

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

$= \frac{- {x}^{3} \sin x - 2 {x}^{2} \cos x + 2 x \sin x}{x} ^ 4 =$

$= \frac{- {x}^{2} \sin x - 2 x \cos x + 2 \sin x}{x} ^ 3$

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

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

$= \frac{- {x}^{3} \cos x + 3 {x}^{4} \sin x + 6 x \cos x - 6 \sin x}{x} ^ 4$.