differentiate the function.\ng(t)=\\ln\\left(\\frac{t(t^{2}+1)^{4}}{\\sqrt9{6t - 1}}\\right)\ng(t)=\\square

Explanation:

Step1: Use logarithm properties

Use $\ln(\frac{a}{b})=\ln a-\ln b$ and $\ln(ab)=\ln a+\ln b$, $\ln(a^n)=n\ln a$. $$g(t)=\ln(t(t^{2}+1)^{4})-\ln((6t - 1)^{\frac{1}{9}})=\ln t+4\ln(t^{2}+1)-\frac{1}{9}\ln(6t - 1)$$

Step2: Differentiate term - by - term

Differentiate using $\frac{d}{dt}(\ln u)=\frac{u'}{u}$. For $\ln t$: $\frac{d}{dt}(\ln t)=\frac{1}{t}$. For $4\ln(t^{2}+1)$: Let $u = t^{2}+1$, then $\frac{d}{dt}(4\ln(t^{2}+1))=4\times\frac{2t}{t^{2}+1}=\frac{8t}{t^{2}+1}$. For $-\frac{1}{9}\ln(6t - 1)$: Let $u = 6t - 1$, then $\frac{d}{dt}(-\frac{1}{9}\ln(6t - 1))=-\frac{1}{9}\times\frac{6}{6t - 1}=-\frac{2}{3(6t - 1)}$.

Step3: Combine the derivatives

$$g'(t)=\frac{1}{t}+\frac{8t}{t^{2}+1}-\frac{2}{3(6t - 1)}$$

Answer:

$$\frac{1}{t}+\frac{8t}{t^{2}+1}-\frac{2}{3(6t - 1)}$$