# What is the derivative of (xe^-x)/(x^3+x)?

Jun 6, 2015

Let's use product rule, here, shall we?

$f \left(x\right) = \left(x {e}^{-} x\right) {\left({x}^{3} + x\right)}^{-} 1$

We'll also need product rule to find the first term's derivative and chain rule for the second one's.

Rules:

• Product rule: be $y = f \left(x\right) g \left(x\right)$, then $y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$
• Chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{-} x \left(1 - x\right) \left({x}^{3} + x\right) + x {e}^{-} x \left(- 1\right) \left(3 {x}^{2} + 1\right)$

#(dy)/(dx)=e^-x(-x^4-x^2-2x^3)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - x\right) \left({x}^{3} + x\right)}{e} ^ x - \frac{3 {x}^{3} + x}{e} ^ x = - \frac{{x}^{4} + 2 {x}^{3} + {x}^{2}}{e} ^ x = \frac{{x}^{2} \left({x}^{2} + 2 x + 1\right)}{e} ^ x = \textcolor{g r e e n}{\frac{{x}^{2} {\left(x + 1\right)}^{2}}{e} ^ x}$