find f(x).\nf(x)=\frac{4x - 7}{6x + 1}\nf(x)=square

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}$. Here, $u(x) = 4x - 7$ and $v(x)=6x + 1$.

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

Differentiate $u(x)$ and $v(x)$ with respect to $x$. Since $u(x)=4x - 7$, then $u^{\prime}(x)=4$. Since $v(x)=6x + 1$, then $v^{\prime}(x)=6$.

Step3: Apply the quotient - rule

Substitute $u(x),u^{\prime}(x),v(x),v^{\prime}(x)$ into the quotient - rule formula. $f^{\prime}(x)=\frac{4(6x + 1)-(4x - 7)\times6}{(6x + 1)^2}$.

Step4: Expand and simplify

Expand the numerator: [ \begin{align*} 4(6x + 1)-(4x - 7)\times6&=24x+4-(24x-42)\ &=24x + 4-24x + 42\ &=46 \end{align*} ] So, $f^{\prime}(x)=\frac{46}{(6x + 1)^2}$.

Answer:

$\frac{46}{(6x + 1)^2}$