find the derivative and simplify.\n\frac{d}{dx}\frac{ln(2 + x)}{x^{3}}\n\nchoose the correct answer…

find the derivative and simplify.\n\frac{d}{dx}\frac{ln(2 + x)}{x^{3}}\n\nchoose the correct answer below.\n\na. \frac{x - 3(2 + x)ln(2 + x)}{x^{4}(2 + x)}\nb. \frac{1 - 6x^{2}(2 + x)ln(2 + x)}{2x^{3}(2 + x)}\nc. \frac{3(2 + x)ln(2 + x)-x}{x^{4}(2 + x)}\nd. \frac{6x^{2}(2 + x)ln(2 + x)-1}{2x^{3}(2 + x)}

find the derivative and simplify.\n\frac{d}{dx}\frac{ln(2 + x)}{x^{3}}\n\nchoose the correct answer below.\n\na. \frac{x - 3(2 + x)ln(2 + x)}{x^{4}(2 + x)}\nb. \frac{1 - 6x^{2}(2 + x)ln(2 + x)}{2x^{3}(2 + x)}\nc. \frac{3(2 + x)ln(2 + x)-x}{x^{4}(2 + x)}\nd. \frac{6x^{2}(2 + x)ln(2 + x)-1}{2x^{3}(2 + x)}

Answer

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, $u = \ln(2 + x)$ and $v=x^{3}$.

Step2: Find $u^\prime$ and $v^\prime$

The derivative of $u=\ln(2 + x)$ using the chain - rule is $u^\prime=\frac{1}{2 + x}$, and the derivative of $v = x^{3}$ is $v^\prime=3x^{2}$.

Step3: Substitute into quotient - rule

$y^\prime=\frac{\frac{1}{2 + x}\cdot x^{3}-\ln(2 + x)\cdot3x^{2}}{(x^{3})^{2}}$.

Step4: Simplify the expression

[ \begin{align*} y^\prime&=\frac{\frac{x^{3}}{2 + x}-3x^{2}\ln(2 + x)}{x^{6}}\ &=\frac{x^{3}-3x^{2}(2 + x)\ln(2 + x)}{x^{6}(2 + x)}\ &=\frac{x - 3(2 + x)\ln(2 + x)}{x^{4}(2 + x)} \end{align*} ]

Answer:

A. $\frac{x - 3(2 + x)\ln(2 + x)}{x^{4}(2 + x)}$