# Getting the derivative of this time dependent vector.

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 $\vec{\ddot b}(\color{blue}{t})$ ?

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?

• 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 – Jasser Dec 24 '15 at 7:20
• @Jasser. that may seem easy to you. but not for me. and I can't have any mistakes in this thing. – AHB Dec 24 '15 at 7:21
• Have you tried? where were you stuck at? – Jasser Dec 24 '15 at 7:23
• @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. – AHB Dec 24 '15 at 7:26
• @Jasser so turn that into an answer. that solved my problem. – AHB Dec 24 '15 at 7:35