# Derivatives of vector-valued functions Differentiate the following function. r(t)=\langle te^{-t},\ t\ln

Derivatives of vector-valued functions Differentiate the following function.
$$\displaystyle{r}{\left({t}\right)}={\left\langle{t}{e}^{{-{t}}},\ {t}{\ln{{t}}},{t}{\cos{{t}}}\right\rangle}$$

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Opeance1951
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,
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{\ln{{t}}}\right)}={\frac{{{t}}}{{{t}}}}+{1}\times{\ln{{t}}}$$
$$\displaystyle={\ln{{t}}}+{1}$$
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{\cos{{t}}}\right)}={\cos{{t}}}+{t}{\left(-{\sin{{t}}}\right)}$$
$$\displaystyle={\cos{{t}}}-{t}{\sin{{t}}}$$
$$\displaystyle\therefore{r}'{\left({t}\right)}={\left\langle{\left({1}-{t}\right)}{e}^{{-{t}}},\ {\left({1}+{\ln{{t}}}\right)},\ {\left({\cos{{t}}}-{t}{\sin{{t}}}\right)}\right\rangle}$$