Question

attempt 1: 10 attempts remaining. calculate the derivative of the function. $g(x)=(6x^{2}+x + 6)^{-8}$ $g(x)=$

Explanation:

Step1: Identify the outer - inner functions

Let $u = 6x^{2}+x + 6$, then $g(x)=u^{-8}$.

Step2: Differentiate the outer function with respect to $u$

Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we have $\frac{dg}{du}=-8u^{-9}$.

Step3: Differentiate the inner function with respect to $x$

$\frac{du}{dx}=\frac{d}{dx}(6x^{2}+x + 6)=12x + 1$.

Step4: Apply the chain - rule

The chain - rule states that $\frac{dg}{dx}=\frac{dg}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dg}{du}$ and $\frac{du}{dx}$:
$\frac{dg}{dx}=-8u^{-9}\cdot(12x + 1)$.

Step5: Substitute $u = 6x^{2}+x + 6$ back in

$g^{\prime}(x)=-8(6x^{2}+x + 6)^{-9}\cdot(12x + 1)=-\frac{8(12x + 1)}{(6x^{2}+x + 6)^{9}}$.

Answer:

$-\frac{8(12x + 1)}{(6x^{2}+x + 6)^{9}}$