## University Calculus: Early Transcendentals (3rd Edition)

$$y'=\frac{1}{x(3x+2)}$$
$$y=\log_{5}\sqrt{\Big(\frac{7x}{3x+2}\Big)^{\ln5}}=\log_{5}\Big(\frac{7x}{3x+2}\Big)^{\frac{\ln5}{2}}$$ We have $\log_ax^b=b\log_ax$. Therefore, $$y=\frac{\ln5}{2}\log_{5}\Big(\frac{7x}{3x+2}\Big)$$ The derivative of $y$ is $$y'=\frac{\ln5}{2}\Big(\log_{5}\Big(\frac{7x}{3x+2}\Big)\Big)'$$ Recall Theorem 7: $$\frac{d}{dx}\log_au=\frac{1}{u\ln a}\frac{du}{dx}$$ That means: $$y'=\frac{\ln5}{2}\times\frac{1}{\frac{7x}{3x+2}\ln5}\Big(\frac{7x}{3x+2}\Big)'$$ $$y'=\frac{1}{2}\times\frac{1}{\frac{7x}{3x+2}}\times\frac{(7x)'(3x+2)-7x(3x+2)'}{(3x+2)^2}$$ $$y'=\frac{1}{2}\times\frac{3x+2}{7x}\times\frac{7(3x+2)-7x(3)}{(3x+2)^2}=\frac{3x+2}{14x}\times\frac{21x+14-21x}{(3x+2)^2}$$ $$y'=\frac{3x+2}{14x}\times\frac{14}{(3x+2)^2}$$ $$y'=\frac{1}{x(3x+2)}$$