Question

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

Answer 1

`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}`