Question

let (h(x)=f(x)g(x)). if (f(x)=4x^{2}-3x) and (g(x)=ln(3x - 1)), what is (h(x))? select the correct answer below: (h(x)=(3 - 8x)ln(3x - 1)+\frac{12x^{2}-9x}{1 - 3x}), (h(x)=(8x - 3)(3x - 1)+\frac{12x^{2}-9x}{3x - 1}), (h(x)=(8x - 3)ln(3x - 1)+\frac{4x^{2}-3x}{3x - 1}), (h(x)=(8x - 3)ln(3x - 1)+\frac{12x^{2}-9x}{3x - 1}), (h(x)=(8x - 3)ln(3x - 1)-\frac{12x^{2}-9x}{3x - 1})

Explanation:

Step1: Recall product - rule

The product - rule states that if $h(x)=f(x)g(x)$, then $h^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$.

Step2: Find $f^{\prime}(x)$

Given $f(x)=4x^{2}-3x$, using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $f^{\prime}(x)=\frac{d}{dx}(4x^{2}-3x)=8x - 3$.

Step3: Find $g^{\prime}(x)$

Given $g(x)=\ln(3x - 1)$, using the chain - rule $\frac{d}{dx}\ln(u)=\frac{u^{\prime}}{u}$ where $u = 3x-1$ and $u^{\prime}=3$, we get $g^{\prime}(x)=\frac{3}{3x - 1}$.

Step4: Calculate $h^{\prime}(x)$

Substitute $f(x)$, $f^{\prime}(x)$, $g(x)$ and $g^{\prime}(x)$ into the product - rule formula:
$h^{\prime}(x)=(8x - 3)\ln(3x - 1)+(4x^{2}-3x)\frac{3}{3x - 1}=\frac{(8x - 3)(3x - 1)\ln(3x - 1)+3(4x^{2}-3x)}{3x - 1}=(8x - 3)\ln(3x - 1)+\frac{12x^{2}-9x}{3x - 1}$.

Answer:

$h^{\prime}(x)=(8x - 3)\ln(3x - 1)+\frac{12x^{2}-9x}{3x - 1}$ (corresponding to the fourth option in the multiple - choice list)