Question

find f(x).
f(x)=\frac{8x - 1}{7x + 2}
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, $u(x)=8x - 1$ and $v(x)=7x + 2$.

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

Differentiate $u(x)=8x - 1$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax + b)=a$, we get $u^{\prime}(x)=8$. Differentiate $v(x)=7x + 2$ with respect to $x$, we get $v^{\prime}(x)=7$.

Step3: Apply the quotient - rule

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

$$\begin{align*} f^{\prime}(x)&=\frac{8(7x + 2)-(8x - 1)\times7}{(7x + 2)^{2}}\\ &=\frac{56x+16-(56x - 7)}{(7x + 2)^{2}}\\ &=\frac{56x + 16-56x + 7}{(7x + 2)^{2}}\\ &=\frac{23}{(7x + 2)^{2}} \end{align*}$$

\]

Answer:

$\frac{23}{(7x + 2)^{2}}$