find the first derivative of $g(t)=lnleft(\frac{e^{t}}{t^{13}}\right)$. $g(t)=$

Explanation:

Step1: Use log - property

Use $\ln(\frac{a}{b})=\ln(a)-\ln(b)$. So $g(t)=\ln(e^{t})-\ln(t^{13})$. Since $\ln(e^{t}) = t$ and $\ln(t^{13})=13\ln(t)$ by the property $\ln(a^{b}) = b\ln(a)$, then $g(t)=t - 13\ln(t)$.

Step2: Differentiate term - by - term

The derivative of $t$ with respect to $t$ is $1$ (using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, here $n = 1$). The derivative of $13\ln(t)$ with respect to $t$ is $\frac{13}{t}$ (using the rule $\frac{d}{dt}(\ln(t))=\frac{1}{t}$). So $g'(t)=\frac{d}{dt}(t)-\frac{d}{dt}(13\ln(t))=1-\frac{13}{t}$.

Answer:

$1-\frac{13}{t}$