#### Answer

$h'(x)=-\dfrac{x^{2}+2x+2}{x^{3}e^{x}}$

#### Work Step by Step

$h(x)=\dfrac{x+1}{x^{2}e^{x}}$
Start the differentiation process by using the quotient rule:
$h'(x)=\dfrac{(x^{2}e^{x})(x+1)'-(x+1)(x^{2}e^{x})'}{(x^{2}e^{x})^{2}}=...$
$...=\dfrac{(x^{2}e^{x})(x+1)'-(x+1)(x^{2}e^{x})'}{x^{4}e^{2x}}=...$
Evaluate the derivatives indicated in the numerator. Use the product rule to evaluate $(x^{2}e^{x})'$:
$...=\dfrac{(x^{2}e^{x})(1)-(x+1)[x^{2}(e^{x})'+(x^{2})'e^{x}]}{x^{4}e^{2x}}=...$
Evaluate the derivatives indicated and start simplifying:
$...=\dfrac{x^{2}e^{x}-(x+1)(x^{2}e^{x}+2xe^{x})}{x^{4}e^{2x}}=...$
Take out common factor $xe^{x}$ from $(x^{2}e^{x}+2xe^{x})$:
$...=\dfrac{x^{2}e^{x}-xe^{x}(x+1)(x+2)}{x^{4}e^{2x}}=...$
Take out common factor $xe^{x}$ from the numerator and simplify:
$...=\dfrac{xe^{x}[x-(x+1)(x+2)]}{x^{4}e^{2x}}=\dfrac{x-(x+1)(x+2)}{x^{3}e^{x}}=...$
$...=\dfrac{-x^{2}-2x-2}{x^{3}e^{x}}=-\dfrac{x^{2}+2x+2}{x^{3}e^{x}}$