Question

find the derivative and simplify.
\frac{d}{dx}\frac{\ln(4 + x)}{x^{9}}

choose the correct answer below.

a. \frac{1-36x^{8}(4 + x)\ln(4 + x)}{4x^{9}(4 + x)}
b. \frac{36x^{8}(4 + x)\ln(4 + x)-1}{4x^{9}(4 + x)}
c. \frac{x - 9(4 + x)\ln(4 + x)}{x^{10}(4 + x)}
d. \frac{9(4 + x)\ln(4 + x)-x}{x^{10}(4 + x)}

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \ln(4 + x)$ and $v=x^{9}$. First, find $u^\prime$ and $v^\prime$. The derivative of $\ln(4 + x)$ with respect to $x$ is $\frac{1}{4 + x}$ (by the chain - rule, since the derivative of $\ln(u)$ with $u = 4 + x$ is $\frac{1}{u}\cdot u^\prime$ and $u^\prime=1$), and the derivative of $v=x^{9}$ with respect to $x$ is $v^\prime = 9x^{8}$.

Step2: Substitute into quotient - rule

$y^\prime=\frac{\frac{1}{4 + x}\cdot x^{9}-\ln(4 + x)\cdot9x^{8}}{(x^{9})^{2}}$.

Step3: Simplify the expression

Multiply the numerator and denominator by $4 + x$ to get rid of the fraction in the numerator:
\[

$$\begin{align*} y^\prime&=\frac{x^{9}-9x^{8}(4 + x)\ln(4 + x)}{x^{18}(4 + x)}\\ &=\frac{x^{8}(x - 9(4 + x)\ln(4 + x))}{x^{18}(4 + x)}\\ &=\frac{x - 9(4 + x)\ln(4 + x)}{x^{10}(4 + x)} \end{align*}$$

\]

Answer:

C. $\frac{x - 9(4 + x)\ln(4 + x)}{x^{10}(4 + x)}$