# Thread: How do I find Dx tan(x)^secx?

1. ## How do I find Dx tan(x)^secx?

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

2. Originally Posted by circuscircus
How would I find the derivative of $\displaystyle tan(x)^{secx}$
we have some variable as a power...logarithmic differentiation should come to mind

3. Or we can use the well-known trick $\displaystyle a=e^{\ln a},\,\forall a>0$

4. Originally Posted by Krizalid
Or we can use the well-known trick $\displaystyle a=e^{\ln a},\,\forall a>0$
right (that trick always slips my mind)

5. I don't see the connection between that and my original equation...

6. Originally Posted by circuscircus
I don't see the connection between that and my original equation...
Krizalid is saying you could notice that $\displaystyle ( \tan x )^{ \sec x} = e^{\sec x \ln ( \tan x)}$. and find the derivative of that. $\displaystyle \left( \frac d{dx}e^u = u'e^u \right)$

7. $\displaystyle e^{\sec x \ln ( \tan x)} secxtanx * \frac{1}{tan x} * sec^2u$

so like this?

8. Originally Posted by circuscircus
$\displaystyle e^{\sec x \ln ( \tan x)} secxtanx * \frac{1}{tan x} * sec^2u$

so like this?
to find the derivative of $\displaystyle \sec x \ln ( \tan x)$ you need to use the product rule (while simultaneously using the chain rule to deal with the $\displaystyle \ln ( \tan x )$ part)

9. $\displaystyle e^{\sec x \ln ( \tan x)} \sec x\tan x \ln ( \tan x) + \sec x \frac{1}{\tan x}\sec^2x$

so would be like this?

10. Originally Posted by circuscircus
$\displaystyle e^{\sec x \ln ( \tan x)} \sec x\tan x \ln ( \tan x) + \sec x \frac{1}{\tan x}\sec^2x$

so would be like this?
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