Question

question let $h(x)=\frac{f(x)}{g(x)}$, where $f(x)=ln(4x^{2}-3x)$ and $g(x)=3x - 1$. what is $h(x)$? select the correct answer below: $h(x)=\frac{\frac{8x - 3}{4x^{2}-3x}(3x - 1)-ln(4x^{2}-3x)(3)}{(3x - 1)^{2}}$ $h(x)=\frac{\frac{8x+3}{4x^{2}-3x}(3x + 1)-ln(4x^{2}+3x)(3)}{(3x - 1)^{2}}$ $h(x)=\frac{\frac{8x - 3}{4x^{2}-3x}(3x - 1)-ln(4x^{2}-3x)(3)}{3x - 1}$ $h(x)=\frac{\frac{8x - 3}{4x^{2}-3x}(3x - 1)-ln(4x^{2}-3x)(3)}{(3x - 1)^{2}}$

Explanation:

Step1: Apply quotient - rule

$h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$
First, find $f'(x)$ and $g'(x)$. For $f(x)=\ln(4x^{2}-3x)$, $f'(x)=\frac{8x - 3}{4x^{2}-3x}$ by chain - rule. For $g(x)=3x - 1$, $g'(x)=3$.

Step2: Substitute into quotient - rule

$h'(x)=\frac{\frac{8x - 3}{4x^{2}-3x}(3x - 1)-\ln(4x^{2}-3x)\times3}{(3x - 1)^{2}}$

Answer:

$h'(x)=\frac{\frac{8x - 3}{4x^{2}-3x}(3x - 1)-3\ln(4x^{2}-3x)}{(3x - 1)^{2}}$