## Calculus: Early Transcendentals (2nd Edition)

$g'(x)=\dfrac{e^{x}(x^{2}-x-5)}{(x-2)^{2}}$
$g(x)=\dfrac{(x+1)e^{x}}{x-2}$ Start the differentiation process by using the quotient rule: $g'(x)=\dfrac{(x-2)[(x+1)e^{x}]'-(x-2)'(x+1)e^{x}}{(x-2)^{2}}=...$ Evaluate the derivatives indicated in the numerator and simplify. Use the product rule to evaluate $[(x+1)e^{x}]'$: $...=\dfrac{(x-2)[(x+1)'e^{x}+(x+1)(e^{x})']-(1)(x+1)e^{x}}{(x-2)^{2}}=...$ $...=\dfrac{(x-2)[(1)e^{x}+e^{x}(x+1)]-(x+1)e^{x}}{(x-2)^{2}}=...$ $...=\dfrac{(x-2)[e^{x}+e^{x}(x+1)]-e^{x}(x+1)}{(x-2)^{2}}=...$ $...=\dfrac{e^{x}(x-2)(1+x+1)-e^{x}(x+1)}{(x-2)^{2}}=...$ $...=\dfrac{e^{x}[(x-2)(x+2)-(x+1)]}{(x-2)^{2}}=...$ $...=\dfrac{e^{x}(x^{2}-4-x-1)}{(x-2)^{2}}=\dfrac{e^{x}(x^{2}-x-5)}{(x-2)^{2}}$