$$\text { Find the derivative of } y \text { with respect to } x , t , \text { or } \theta \text {, as appropriate. }$$
-$$y = \ln \frac { 1 + \sqrt { x } } { x ^ { 3 } }$$
A) $\frac { - 6 - 5 \sqrt { x } } { 2 ( 1 + \sqrt { x } ) }$
B) $\frac { 6 - 5 \sqrt { x } } { 2 x ( 1 + \sqrt { x } ) }$
C) $\frac { - 6 - 5 \sqrt { x } } { 2 x ( 1 + \sqrt { x } ) }$
D) $\frac { - 6 - 5 \sqrt { x } } { 2 x }$

Question

Quizplus · Accepted Answer

The derivative of $y = \ln \frac { 1 + \sqrt { x } } { x ^ { 3 } }$ can be found using the chain rule and the properties of logarithms. First, rewrite the expression using logarithm properties: $y = \ln(1 + \sqrt{x}) - \ln(x^3)$. The derivative of $\ln(1 + \sqrt{x})$ with respect to $x$ is $\frac{1}{1 + \sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$, and the derivative of $\ln(x^3)$ is $\frac{3}{x}$. Combining these gives $\frac{1}{2\sqrt{x}(1 + \sqrt{x})} - \frac{3}{x}$, which simplifies to $\frac{-6 - 5\sqrt{x}}{2x(1 + \sqrt{x})}$.

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.

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