How would I find the derivative of $\displaystyle tan(x)^{secx}$

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- Sep 25th 2007, 10:44 AMcircuscircusHow do I find Dx tan(x)^secx?
How would I find the derivative of $\displaystyle tan(x)^{secx}$

- Sep 25th 2007, 11:01 AMJhevon
- Sep 25th 2007, 11:50 AMKrizalid
Or we can use the well-known trick $\displaystyle a=e^{\ln a},\,\forall a>0$

- Sep 25th 2007, 11:57 AMJhevon
- Sep 25th 2007, 09:32 PMcircuscircus
I don't see the connection between that and my original equation...

- Sep 25th 2007, 09:54 PMJhevon
- Sep 25th 2007, 10:40 PMcircuscircus
$\displaystyle e^{\sec x \ln ( \tan x)} secxtanx * \frac{1}{tan x} * sec^2u

$

so like this? - Sep 25th 2007, 10:43 PMJhevon
- Sep 25th 2007, 11:07 PMcircuscircus
$\displaystyle e^{\sec x \ln ( \tan x)} \sec x\tan x \ln ( \tan x) + \sec x \frac{1}{\tan x}\sec^2x$

so would be like this? - Sep 25th 2007, 11:27 PMJhevon
use parentheses! you have the idea, but the answer as written is wrong.

you should have: $\displaystyle \left( \sec x\tan x \ln ( \tan x) + \sec x \frac{1}{\tan x}\sec^2x \right)e^{\sec x \ln ( \tan x)}$

by the way, this can be simplified a lot, for instance, instead of writing $\displaystyle \sec x \frac 1{\tan x} \sec^2 x$ you could write $\displaystyle \frac {\sec^3 x}{\tan x}$ and you can change the $\displaystyle e^{\sec x \ln \tan x}$ back to it's original form