Question

find f(x).
f(x) = \frac{x^{2}+7}{3x - 1}
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, $u(x)=x^{2}+7$ and $v(x)=3x - 1$.

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

Differentiate $u(x)$: $u^{\prime}(x)=\frac{d}{dx}(x^{2}+7)=2x$. Differentiate $v(x)$: $v^{\prime}(x)=\frac{d}{dx}(3x - 1)=3$.

Step3: Apply the quotient - rule

$f^{\prime}(x)=\frac{2x(3x - 1)-(x^{2}+7)\times3}{(3x - 1)^{2}}$.

Step4: Expand the numerator

Expand $2x(3x - 1)-(x^{2}+7)\times3$:
\[

$$\begin{align*} 2x(3x - 1)-(x^{2}+7)\times3&=6x^{2}-2x-(3x^{2}+21)\\ &=6x^{2}-2x - 3x^{2}-21\\ &=3x^{2}-2x - 21 \end{align*}$$

\]
So, $f^{\prime}(x)=\frac{3x^{2}-2x - 21}{(3x - 1)^{2}}$.

Answer:

$\frac{3x^{2}-2x - 21}{(3x - 1)^{2}}$