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This vector is time dependent with the following formula:

$$ \begin{align} \vec b(\color{blue}{t}) = & \ ( \color{#e69900}{r} \cos(\color{#e69900}{\omega} \color{blue}{t})+\color{#e69900}{l}\sin \color{blue}{\theta} \cos \color{blue}{\phi})\color{#e69900}{\hat i} \\ & +(\color{#e69900}{r}\sin(\color{#e69900}{\omega} \color{blue}{t})+\color{#e69900}{l}\sin \color{blue}{\theta} \sin \color{blue}{\phi})\color{#e69900}{\hat j} \\ & -\color{#e69900}{l}\cos \color{blue}{\theta}\ \color{#e69900}{\hat k} \end{align}$$

$\begin{align} \text{I have colored } & \text{time dependents } \color{blue}{\text{blue}} \text{.} \\ & \text{and constants } \color{#e69900}{\text{golden}} \text{.} \\ \end{align}$

How can I calculate $\vec{\dot b}(\color{blue}{t})$ and $\vec{\ddot b}(\color{blue}{t})$ ?

I know basic derivative rules, but this one is hard for me. please help.

My main problem is with taking the derivative of $\sin \color{blue}{\theta} \cos \color{blue}{\phi}$. Should I apply the chain rule or the product rule first?

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  • $\begingroup$ derivative of a vector is the sum of derivatives of its components along the given axes.... and to go on you can even use product rule $\endgroup$ – Jasser Dec 24 '15 at 7:20
  • $\begingroup$ @Jasser. that may seem easy to you. but not for me. and I can't have any mistakes in this thing. $\endgroup$ – AHB Dec 24 '15 at 7:21
  • $\begingroup$ Have you tried? where were you stuck at? $\endgroup$ – Jasser Dec 24 '15 at 7:23
  • $\begingroup$ @Jesser. when taking the derivative of $\sin(\theta)\cos(\phi)$, it's a product and should obey product rule. also it has chain functions, so it should obey it ,too. which rule to use first now ?confused. $\endgroup$ – AHB Dec 24 '15 at 7:26
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    $\begingroup$ @Jasser so turn that into an answer. that solved my problem. $\endgroup$ – AHB Dec 24 '15 at 7:35
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As I said in the comment derivative of a vector is the sum of derivatives of its components along the given axes so take the derivatives

Then use product rule and you will see the chain rule emerging out.

Proceed in a similar way to find out the second derivative...

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