Answer
$$\frac{d y}{d x} =\frac{1}{x(3 x+2)}$$
Work Step by Step
Given $$ y =\log _{5} \sqrt{\left(\frac{7 x}{3 x+2}\right)^{\ln 5}}$$
Since $$\log_{a}z=\frac{\ln z}{\ln a}$$ So, we have
\begin{aligned}
y&=\log _{5}\left(\frac{7 x}{3 x+2}\right)^{(1 n 5) / 2}\\
&=\frac{\ln \left(\frac{7 x}{3 x+2}\right)^{(\ln 5) / 2}}{\ln 5}\\
&=\left(\frac{\ln 5}{2}\right)\left[\frac{\ln \left(\frac{7 x}{3 x+2}\right)}{\ln 5}\right]\\
&=\frac{1}{2} \ln \left(\frac{7 x}{3 x+2}\right)\\
&=\frac{1}{2} \ln 7 x-\frac{1}{2} \ln (3 x+2) \\
&\Rightarrow \frac{d y}{d x}=\frac{7}{2 \cdot 7 x}-\frac{3}{2 \cdot(3 x+2)}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2 x}-\frac{3}{2 (3 x+2)}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{(3 x+2)-3 x}{2 x(3 x+2)}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2}{2x(3 x+2)}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{x(3 x+2)}\\
\end{aligned}
