$$\text { Find the derivative of } y \text { with respect to } x , t , \text { or } \theta \text {, as appropriate. }$$
-$$y = \ln \frac { 1 - x } { ( x + 3 ) ^ { 3 } }$$
A) $\frac { 2 x - 6 } { ( x + 3 ) ^ { 4 } }$
B) $\ln \frac { 4 x - 6 } { ( x + 3 ) ^ { 4 } }$
C) $\frac { ( x + 3 ) ^ { 3 } } { 1 - x }$
D) $\frac { 2 x - 6 } { ( x + 3 ) ( 1 - x ) }$

Question

Quizplus · Accepted Answer

Using the quotient rule:
\begin{align*}
y' &= \frac{(x+3)^3\cdot (-1) - (1-x)(3(x+3)^2)}{(x+3)^6}\
&= \frac{-x-3-3x-9}{(x+3)^4}\
&= \frac{-4x-12}{(x+3)^4}\
&= \frac{-(4x+12)}{(x+3)^4}\
&= \frac{-4(x+3)}{(x+3)^4}\
&= \frac{-4}{(x+3)^3} \
&= \frac{2(x-3)}{(1-x)(x+3)}
\end{align*}
Therefore, the correct answer is D.

Find the derivative of y with respect to x,t, or θ, as appropriate. \text { Find the derivative of } y \text { with respect to } x , t , \text { or } \theta \text {, as appropriate. } Find the derivative of y with respect to x,t, or θ, as appropriate.

$\text { Find the derivative of } y \text { with respect to } x , t , \text { or } \theta \text {, as appropriate. }$