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

Answer 1

Samantha Adkison

#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}#

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Answer 2

Christopher Allers

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.