Question

# Find the derivatives of the functions. r(t)=\log_3(t+\sqrt t)

Derivatives
Find the derivatives of the functions.
$$\displaystyle{r}{\left({t}\right)}={{\log}_{{3}}{\left({t}+\sqrt{{t}}\right)}}$$

2021-06-24
$$\displaystyle{r}{\left({t}\right)}={{\log}_{{3}}{\left({t}+\sqrt{{t}}\right)}}$$
Rewrite the function, recall that $$\displaystyle\sqrt{{a}}={a}^{{{\frac{{{1}}}{{{2}}}}}}$$, so
$$\displaystyle{r}{\left({t}\right)}={{\log}_{{3}}{\left({t}+{t}^{{{\frac{{{1}}}{{{2}}}}}}\right)}}$$
Diffetrentiate both sides with respect to x
$$\displaystyle{r}'{\left({t}\right)}={\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left[{{\log}_{{3}}{\left({t}+{t}^{{{\frac{{{1}}}{{{2}}}}}}\right)}}\right]}$$
Apply $$\displaystyle{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left[{{\log}_{{b}}^{{\left|{u}\right|}}}\right]}={\frac{{{1}}}{{{u}{\ln{{b}}}}}}{\left\lbrace{\frac{{{d}{u}}}{{{\left.{d}{t}\right.}}}}\right.}$$, let $$\displaystyle{u}={t}+{t}^{{{\frac{{{1}}}{{{2}}}}}}$$, so
$$\displaystyle{r}'{\left({t}\right)}={\frac{{{1}}}{{{\left({t}+{t}^{{{\frac{{{1}}}{{{2}}}}}}\right)}{\ln{{3}}}}}}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left[{t}+{t}^{{{\frac{{{1}}}{{{2}}}}}}\right]}$$
Therefore,
$$\displaystyle{r}'{\left({t}\right)}={\frac{{{1}}}{{{\left({t}+{t}^{{{\frac{{{1}}}{{{2}}}}}}\right)}{\ln{{3}}}}}}{\left({1}+{\frac{{{1}}}{{{2}}}}{t}^{{-{\frac{{{1}}}{{{2}}}}}}\right)}$$
Simplify
$$\displaystyle{r}'{\left({t}\right)}={\frac{{{1}+{\frac{{{1}}}{{{2}}}}{t}^{{-{\frac{{{1}}}{{{2}}}}}}}}{{{\left({t}+{t}^{{{\frac{{{1}}}{{{2}}}}}}\right)}{\ln{{3}}}}}}$$
or
$$\displaystyle{r}'{\left({t}\right)}={\frac{{{1}+{\frac{{{1}}}{{{2}\sqrt{{t}}}}}}}{{{\left({t}+\sqrt{{t}}\right)}{\ln{{3}}}}}}$$