find f(x).\nf(x)=\frac{4x - 7}{9x + 1}\nf(x)=square

Explanation:

Step1: Apply 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, $u(x)=4x - 7$, $u^{\prime}(x)=4$, $v(x)=9x + 1$, and $v^{\prime}(x)=9$.

Step2: Substitute into the formula

$f^{\prime}(x)=\frac{4(9x + 1)-(4x - 7)\times9}{(9x + 1)^2}$.

Step3: Expand the numerator

First, expand $4(9x + 1)=36x+4$ and $(4x - 7)\times9 = 36x-63$. Then the numerator is $36x + 4-(36x - 63)=36x + 4-36x+63$.

Step4: Simplify the numerator and the derivative

The numerator simplifies to $67$. So, $f^{\prime}(x)=\frac{67}{(9x + 1)^2}$.

Answer:

$\frac{67}{(9x + 1)^2}$