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

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To find the derivative of ( \frac{xe^{-x}}{x^3+x} ), apply the quotient rule.

Let ( u = xe^{-x} ) and ( v = x^3 + x ).

Using the quotient rule:

( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} )

Where ( u' ) and ( v' ) represent the derivatives of ( u ) and ( v ) respectively.

( u' = e^{-x} - xe^{-x} )

( v' = 3x^2 + 1 )

Plugging these into the formula:

( \frac{d}{dx}\left(\frac{xe^{-x}}{x^3+x}\right) = \frac{(e^{-x} - xe^{-x})(x^3+x) - (xe^{-x})(3x^2+1)}{(x^3+x)^2} )

Simplify this expression for the derivative.

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