cleritere39

2021-11-27

Compute the derivative of the vector-valued function $r\left(t\right)=\u27e8\mathrm{sin}2t,e{t}^{2}\u27e9.$

George Woody

Beginner2021-11-28Added 16 answers

Step 1

In order to compute the derivative of the vector-valued function:

Differentiate the vector function on a component-by-component basis using the chain rule

Step 2

We differentiate the vector function on a component-by-component basis using the chain rule. This yields:

${\left(\mathrm{sin}2t\right)}^{{}^{\prime}}=\mathrm{cos}2t\cdot {\left(2t\right)}^{{}^{\prime}}=2\mathrm{cos}2t$

$\left({e}^{{t}^{2}}\right)}^{{}^{\prime}}={e}^{{t}^{2}}\cdot {\left({t}^{2}\right)}^{{}^{\prime}}=2t{e}^{{t}^{2}$ .

So the derivative is given by

${r}^{{}^{\prime}}\left(t\right)=\u27e82\mathrm{cos}2t,2t{e}^{{t}^{2}}\u27e9.$

Step 3

Hence, the final answer is:

${r}^{{}^{\prime}}\left(t\right)=\u27e82\mathrm{cos}2t,2t{e}^{{t}^{2}}\u27e9.$

In order to compute the derivative of the vector-valued function:

Differentiate the vector function on a component-by-component basis using the chain rule

Step 2

We differentiate the vector function on a component-by-component basis using the chain rule. This yields:

So the derivative is given by

Step 3

Hence, the final answer is: