question use the quotient rule to find the derivative, h(x), of the following function. h(x)=\frac{2x}{-8x…

Explanation:

Step1: Recall quotient - rule formula

The quotient - rule states that if $h(x)=\frac{u(x)}{v(x)}$, then $h^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}$. Here, $u(x)=2x$ and $v(x)= - 8x - 1$.

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

The derivative of $u(x)=2x$ is $u^{\prime}(x) = 2$ (using the power - rule $\frac{d}{dx}(ax)=a$ where $a = 2$). The derivative of $v(x)=-8x - 1$ is $v^{\prime}(x)=-8$ (since $\frac{d}{dx}(-8x)=-8$ and $\frac{d}{dx}(-1)=0$).

Step3: Apply the quotient - rule

Substitute $u(x)$, $v(x)$, $u^{\prime}(x)$, and $v^{\prime}(x)$ into the quotient - rule formula: [ \begin{align*} h^{\prime}(x)&=\frac{2(-8x - 1)-2x(-8)}{(-8x - 1)^2}\ &=\frac{-16x-2 + 16x}{(-8x - 1)^2}\ &=\frac{-2}{(-8x - 1)^2} \end{align*} ]

Answer:

$\frac{-2}{(-8x - 1)^2}$