Question

differentiate the function.

$g(y)=lnleft(\frac{(2y + 1)^3}{sqrt{y^2 + 1}}
ight)$

$g(y)=$

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:

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