Question

find f(x).
f(x)=\frac{x^{2}+6}{9x - 8}
f(x)=square

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x)=x^{2}+6$ and $v(x)=9x - 8$.

Step2: Find $u^{\prime}(x)$ and $v^{\prime}(x)$

Differentiate $u(x)=x^{2}+6$ using the power - rule. The derivative of $x^{n}$ is $nx^{n - 1}$, so $u^{\prime}(x)=2x$. Differentiate $v(x)=9x - 8$, then $v^{\prime}(x)=9$.

Step3: Apply the quotient - rule

Substitute $u(x),u^{\prime}(x),v(x),v^{\prime}(x)$ into the quotient - rule formula:
\[

$$\begin{align*} f^{\prime}(x)&=\frac{(2x)(9x - 8)-(x^{2}+6)\times9}{(9x - 8)^{2}}\\ &=\frac{18x^{2}-16x-(9x^{2}+54)}{(9x - 8)^{2}}\\ &=\frac{18x^{2}-16x - 9x^{2}-54}{(9x - 8)^{2}}\\ &=\frac{9x^{2}-16x - 54}{(9x - 8)^{2}} \end{align*}$$

\]

Answer:

$\frac{9x^{2}-16x - 54}{(9x - 8)^{2}}$