Question

question: use the quotient rule to find the derivative, h(x), of the following function. h(x) = 6x / (2x^2 + 9x - 6). sorry, thats incorrect. try again? enter an exact answer. h(x) =

Explanation:

Step1: Recall quotient - rule formula

The quotient - rule states that if $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g(x)^2}$. Here, $f(x) = 6x$ and $g(x)=2x^{2}+9x - 6$.

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

The derivative of $f(x)=6x$ is $f^{\prime}(x) = 6$ (using the power - rule $\frac{d}{dx}(ax)=a$ for a constant $a$). The derivative of $g(x)=2x^{2}+9x - 6$ is $g^{\prime}(x)=4x + 9$ (using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$).

Step3: Apply the quotient - rule

\[

$$\begin{align*} h^{\prime}(x)&=\frac{6(2x^{2}+9x - 6)-6x(4x + 9)}{(2x^{2}+9x - 6)^{2}}\\ &=\frac{12x^{2}+54x-36-(24x^{2}+54x)}{(2x^{2}+9x - 6)^{2}}\\ &=\frac{12x^{2}+54x - 36-24x^{2}-54x}{(2x^{2}+9x - 6)^{2}}\\ &=\frac{- 12x^{2}-36}{(2x^{2}+9x - 6)^{2}} \end{align*}$$

\]

Answer:

$h^{\prime}(x)=\frac{-12x^{2}-36}{(2x^{2}+9x - 6)^{2}}$