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

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Answer 1 · 2016-06-01T17:05:53

$frac{d}{dx}(frac{xe^{-x}}{x^3+x})=frac{-e^{-x}x^4-e^{-x}x^2-2e^{-x}x^3}{(x^3+x)^2}$

$frac{d}{dx}(frac{xe^{-x}}{x^3+x})$

Applying quotient rule,
$(frac{f}{g})^'=frac{f^'cdot g-g^'\cdot f}{g^2}$

$=frac{frac{d}{dx}(xe^{-x})(x^3+x)-frac{d}{dx}(x^3+x)xe^{-x}}{(x^3+x)^2}$

We know,
$frac{d}{dx}(xe^{-x})=e^{-x}-e^{-x}x$
and,
$frac{d}{dx}(x^3+x)=3x^2+1$

So,
$=frac{(e^{-x}-e^{-x}x)(x^3+x)-(3x^2+1)xe^{-x}}{(x^3+x)^2}$

Simplifying it,we get,
$frac{d}{dx}(frac{xe^{-x}}{x^3+x})=frac{-e^{-x}x^4-e^{-x}x^2-2e^{-x}x^3}{(x^3+x)^2}$

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