given the function $f(x)=(4x^{2}+3x - 7)^{-3}$, find $f(x)$ in any form.\nanswer attempt 1 out of 2\n$f(x)=$

Explanation:

Step1: Identify the outer - inner functions

Let $u = 4x^{2}+3x - 7$, then $y = u^{-3}$.

Step2: Differentiate the outer function

Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, for $y = u^{-3}$, we have $\frac{dy}{du}=-3u^{-4}$.

Step3: Differentiate the inner function

$\frac{du}{dx}=\frac{d}{dx}(4x^{2}+3x - 7)=8x + 3$.

Step4: Apply the chain - rule

The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substituting $\frac{dy}{du}$ and $\frac{du}{dx}$ we found above, we get $\frac{dy}{dx}=-3u^{-4}\cdot(8x + 3)$.

Step5: Substitute $u$ back in

Since $u = 4x^{2}+3x - 7$, then $f^{\prime}(x)=-3(4x^{2}+3x - 7)^{-4}\cdot(8x + 3)=-\frac{3(8x + 3)}{(4x^{2}+3x - 7)^{4}}$.

Answer:

$-\frac{3(8x + 3)}{(4x^{2}+3x - 7)^{4}}$