## Calculus: Early Transcendentals (2nd Edition)

$h'(x)=\dfrac{2x^{2}+6x+1}{(x^{2}+x+1)^{2}}$
$h(x)=\dfrac{(x-1)(2x^{2}-1)}{x^{3}-1}$ Start the differentiation process by using the quotient rule: $h'(x)=\dfrac{(x^{3}-1)[(x-1)(2x^{2}-1)]'-(x^{3}-1)'(x-1)(2x^{2}-1)}{(x^{3}-1)^{2}}=...$ Evaluate the derivatives indicated in the numerator and simplify. Use the product rule to evaluate $[(x-1)(2x^{2}-1)]'$: $...=\dfrac{(x^{3}-1)[(x-1)(2x^{2}-1)'+(2x^{2}-1)(x-1)']-(3x^{2})(x-1)(2x^{2}-1)}{(x^{3}-1)^{2}}=...$ $...=\dfrac{(x^{3}-1)[(x-1)(4x)+(2x^{2}-1)(1)]-(3x^{2})(x-1)(2x^{2}-1)}{(x^{3}-1)^{2}}=...$ $...=\dfrac{(x^{3}-1)[4x^{2}-4x+2x^{2}-1]-3x^{2}(x-1)(2x^{2}-1)}{(x^{3}-1)^{2}}=...$ $...=\dfrac{(x^{3}-1)(6x^{2}-4x-1)-3x^{2}(x-1)(2x^{2}-1)}{(x^{3}-1)^{2}}=...$ $...=\dfrac{6x^{5}-4x^{4}-x^{3}-6x^{2}+4x+1-6x^{5}+6x^{4}+3x^{3}-3x^{2}}{(x^{3}-1)^{2}}=...$ $...=\dfrac{2x^{4}+2x^{3}-9x^{2}+4x+1}{(x-1)^{2}(x^{2}+x+1)^{2}}=...$ $...=\dfrac{(2x^{2}+6x+1)(x-1)^{2}}{(x-1)^{2}(x^{2}+x+1)^{2}}=\dfrac{2x^{2}+6x+1}{(x^{2}+x+1)^{2}}$