#### Answer

$h'(x)=\dfrac{e^{x}(x^{2}+x+1)}{(x+1)^{2}}$

#### Work Step by Step

$h(x)=\dfrac{xe^{x}}{x+1}$
Start the differentiation process by using the quotient rule:
$h'(x)=\dfrac{(x+1)(xe^{x})'-(xe^{x})(x+1)'}{(x+1)^{2}}=...$
Evaluate the derivatives indicated in the numerator and simplify. Use the product rule to evaluate $(xe^{x})'$:
$...=\dfrac{(x+1)[x(e^{x})'+(x)'e^{x}]-xe^{x}(1)}{(x+1)^{2}}=...$
$...=\dfrac{(x+1)(xe^{x}+e^{x})-xe^{x}}{(x+1)^{2}}=...$
Take out common factor $e^{x}$ from $(xe^{x}+e^{x})$:
$...=\dfrac{e^{x}(x+1)(x+1)-xe^{x}}{(x+1)^{2}}=\dfrac{e^{x}(x+1)^{2}-xe^{x}}{(x+1)^{2}}=...$
Take out common factor $e^{x}$ from the numerator:
$...=\dfrac{e^{x}[(x+1)^{2}-x]}{(x+1)^{2}}=...$
Simplify again:
$...=\dfrac{e^{x}(x^{2}+2x+1-x)}{(x+1)^{2}}=\dfrac{e^{x}(x^{2}+x+1)}{(x+1)^{2}}$