Question

differentiate the following function. you do not need to simplify the derivative.

$f(x)=ln(-8x^{2}-6x + 2)$

Explanation:

Step1: Apply chain - rule

Let $u=-8x^{2}-6x + 2$, then $f(x)=\ln(u)$. The chain - rule states that $\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$.

Step2: Differentiate $\ln(u)$ with respect to $u$

The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$. So, $\frac{df}{du}=\frac{1}{u}$.

Step3: Differentiate $u$ with respect to $x$

$u=-8x^{2}-6x + 2$, then $\frac{du}{dx}=-16x-6$.

Step4: Calculate $\frac{df}{dx}$

Substitute $\frac{df}{du}$ and $\frac{du}{dx}$ into the chain - rule formula: $\frac{df}{dx}=\frac{1}{u}\cdot(-16x - 6)$. Replace $u=-8x^{2}-6x + 2$ back in, we get $\frac{df}{dx}=\frac{-16x - 6}{-8x^{2}-6x + 2}$.

Answer:

$\frac{-16x - 6}{-8x^{2}-6x + 2}$