differentiate the function.\n\n$g(y)=lnleft(\frac{(2y + 1)^3}{sqrt{y^2 + 1}}\right)$\n\n$g(y)=$

Explanation:

Step1: Apply logarithm properties

Use (\ln\frac{a}{b}=\ln a-\ln b) and (\ln a^n = n\ln a). (G(y)=\ln(2y + 1)^3-\ln(y^2 + 1)^{\frac{1}{2}}=3\ln(2y + 1)-\frac{1}{2}\ln(y^2 + 1))

Step2: Differentiate term - by - term

Use the chain rule ((\ln u)^\prime=\frac{u^\prime}{u}). For (u = 2y+1), ((3\ln(2y + 1))^\prime=3\times\frac{2}{2y + 1}=\frac{6}{2y + 1}). For (u=y^2 + 1), ((-\frac{1}{2}\ln(y^2 + 1))^\prime=-\frac{1}{2}\times\frac{2y}{y^2 + 1}=-\frac{y}{y^2 + 1}).

Step3: Combine the results

(G^\prime(y)=\frac{6}{2y + 1}-\frac{y}{y^2 + 1})

Answer:

(\frac{6}{2y + 1}-\frac{y}{y^2 + 1})