Question

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

Derivatives
ANSWERED
asked 2021-06-23
Find the derivatives of the functions.
\(\displaystyle{r}{\left({t}\right)}={{\log}_{{3}}{\left({t}+\sqrt{{t}}\right)}}\)

Answers (1)

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
\(\displaystyle{r}'{\left({t}\right)}={\frac{{{1}+{\frac{{{1}}}{{{2}\sqrt{{t}}}}}}}{{{\left({t}+\sqrt{{t}}\right)}{\ln{{3}}}}}}\)
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