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

1 Answer
Jun 1, 2016

Answer:

#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}#

Explanation:

#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}#