Question

find the derivative of the following function.
h(x)=\frac{(x - 1)(9x^{2}-1)}{x^{3}-1}
h(x)=square

Explanation:

Step1: Simplify the function

First, factor \(x^{3}-1=(x - 1)(x^{2}+x + 1)\) using the formula \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\). Then \(h(x)=\frac{(x - 1)(9x^{2}-1)}{(x - 1)(x^{2}+x + 1)}=\frac{9x^{2}-1}{x^{2}+x + 1}\) for \(x
eq1\).

Step2: Apply the quotient - rule

The quotient - rule states that if \(y=\frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here, \(u = 9x^{2}-1\), so \(u^\prime=18x\), and \(v=x^{2}+x + 1\), so \(v^\prime=2x + 1\).
\[

$$\begin{align*} h^\prime(x)&=\frac{(18x)(x^{2}+x + 1)-(9x^{2}-1)(2x + 1)}{(x^{2}+x + 1)^{2}}\\ &=\frac{18x^{3}+18x^{2}+18x-(18x^{3}+9x^{2}-2x - 1)}{(x^{2}+x + 1)^{2}}\\ &=\frac{18x^{3}+18x^{2}+18x - 18x^{3}-9x^{2}+2x + 1}{(x^{2}+x + 1)^{2}}\\ &=\frac{(18x^{3}-18x^{3})+(18x^{2}-9x^{2})+(18x + 2x)+1}{(x^{2}+x + 1)^{2}}\\ &=\frac{9x^{2}+20x + 1}{(x^{2}+x + 1)^{2}} \end{align*}$$

\]

Answer:

\(\frac{9x^{2}+20x + 1}{(x^{2}+x + 1)^{2}}\)