Step 1

Given, \(\displaystyle{r}{\left({t}\right)}={\left\langle{t}{e}^{{-{t}}},\ {t}\ {e}{n}{t},\ {t}{\cos{{t}}}\right\rangle}\)

So, obtain differntiate individually

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{e}^{{-{t}}}\right)}={e}^{{-{t}}}+{t}{\left({e}^{{-{t}}}\right)}{\left(-{1}\right)}\)

\(\displaystyle={\left({1}-{t}\right)}{e}^{{-{t}}}\)

Step 2

Similarly,

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{\ln{{t}}}\right)}={\frac{{{t}}}{{{t}}}}+{1}\times{\ln{{t}}}\)

\(\displaystyle={\ln{{t}}}+{1}\)

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{\cos{{t}}}\right)}={\cos{{t}}}+{t}{\left(-{\sin{{t}}}\right)}\)

\(\displaystyle={\cos{{t}}}-{t}{\sin{{t}}}\)

\(\displaystyle\therefore{r}'{\left({t}\right)}={\left\langle{\left({1}-{t}\right)}{e}^{{-{t}}},\ {\left({1}+{\ln{{t}}}\right)},\ {\left({\cos{{t}}}-{t}{\sin{{t}}}\right)}\right\rangle}\)

Given, \(\displaystyle{r}{\left({t}\right)}={\left\langle{t}{e}^{{-{t}}},\ {t}\ {e}{n}{t},\ {t}{\cos{{t}}}\right\rangle}\)

So, obtain differntiate individually

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{e}^{{-{t}}}\right)}={e}^{{-{t}}}+{t}{\left({e}^{{-{t}}}\right)}{\left(-{1}\right)}\)

\(\displaystyle={\left({1}-{t}\right)}{e}^{{-{t}}}\)

Step 2

Similarly,

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{\ln{{t}}}\right)}={\frac{{{t}}}{{{t}}}}+{1}\times{\ln{{t}}}\)

\(\displaystyle={\ln{{t}}}+{1}\)

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{\cos{{t}}}\right)}={\cos{{t}}}+{t}{\left(-{\sin{{t}}}\right)}\)

\(\displaystyle={\cos{{t}}}-{t}{\sin{{t}}}\)

\(\displaystyle\therefore{r}'{\left({t}\right)}={\left\langle{\left({1}-{t}\right)}{e}^{{-{t}}},\ {\left({1}+{\ln{{t}}}\right)},\ {\left({\cos{{t}}}-{t}{\sin{{t}}}\right)}\right\rangle}\)