## Calculus: Early Transcendentals (2nd Edition)

$h'(x)=-\dfrac{x^{2}+2x+2}{x^{3}e^{x}}$
$h(x)=\dfrac{x+1}{x^{2}e^{x}}$ 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: $...=\dfrac{xe^{x}[x-(x+1)(x+2)]}{x^{4}e^{2x}}=\dfrac{x-(x+1)(x+2)}{x^{3}e^{x}}=...$ $...=\dfrac{-x^{2}-2x-2}{x^{3}e^{x}}=-\dfrac{x^{2}+2x+2}{x^{3}e^{x}}$