Question

calculate the derivative of the function.
g(x)=(7x^{2}+x + 7)^{-8}
g(x)=

Explanation:

Step1: Identify the outer - inner functions

Let $u = 7x^{2}+x + 7$, then $g(x)=u^{-8}$.

Step2: Differentiate the outer function

The derivative of $y = u^{-8}$ with respect to $u$ is $\frac{dy}{du}=-8u^{-9}$ using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$.

Step3: Differentiate the inner function

The derivative of $u = 7x^{2}+x + 7$ with respect to $x$ is $\frac{du}{dx}=14x + 1$ using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$.

Step4: Apply the chain - rule

The chain - rule states that $\frac{dg}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$:
\[

$$\begin{align*} g^{\prime}(x)&=-8u^{-9}\cdot(14x + 1)\\ &=-8(7x^{2}+x + 7)^{-9}\cdot(14x + 1) \end{align*}$$

\]

Answer:

$-8(14x + 1)(7x^{2}+x + 7)^{-9}$