3600244's question via email about a derivative

[Prove It]

What is the derivative (with respect to t) of $\displaystyle \begin{align*} 15\log{ \left| \sec{ \left( 9\,t \right) } + \tan{ \left( 9\,t \right) } \right| } \end{align*}$?

$\displaystyle \begin{align*} y &= 15\log{ \left| \sec{ \left( 9\,t \right) } + \tan{ \left( 9\,t \right) } \right| } \\ &= 15 \log{ \left| \frac{1}{\cos{\left( 9\,t \right) } } + \frac{\sin{ \left( 9\,t \right) }}{\cos{ \left( 9\,t \right) }} \right| } \\ &= 15 \log{ \left| \frac{1 + \sin{\left( 9\,t \right) }}{\cos{ \left( 9\,t \right) }} \right| } \\ &= 15 \left[ \log{\left| 1 + \sin{ \left( 9\,t \right) } \right| } - \log{\left| \cos{ \left( 9\,t \right) } \right| } \right] \end{align*}$

Now differentiating each piece using the rule $\displaystyle \begin{align*} \left( \log{ \left| f(x) \right| } \right) ' = \frac{f'(x)}{f(x)} \end{align*}$ we must have

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= 15 \left[ \frac{9\cos{ \left( 9\,t \right) }}{1 + \sin{ \left( 9\,t \right) }} - \frac{-9\sin{ \left( 9\,t \right) }}{\cos{ \left( 9\,t \right) }} \right] \\ &= 135 \left[ \frac{\cos{\left( 9\,t \right) }}{1 + \sin{ \left( 9\,t \right) }} + \tan{ \left( 9\,t \right) } \right] \end{align*}$

 

[jvicens]

The derivative with respect to t of the given function is $\displaystyle 135 \left[ \frac{\cos{\left( 9\,t \right) }}{1 + \sin{ \left( 9\,t \right) }} + \tan{ \left( 9\,t \right) } \right]$. This can also be written as $\displaystyle \frac{135}{\cos{ \left( 9\,t \right) }}$.