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

**circuscircus** · Sep 25th 2007, 10:44 AM

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

**Jhevon** · Sep 25th 2007, 11:01 AM

Originally Posted by circuscircus

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

we have some variable as a power...logarithmic differentiation should come to mind

**Krizalid** · Sep 25th 2007, 11:50 AM

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

**Jhevon** · Sep 25th 2007, 11:57 AM

Originally Posted by Krizalid

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

right

(that trick always slips my mind)

**circuscircus** · Sep 25th 2007, 09:32 PM

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

**Jhevon** · Sep 25th 2007, 09:54 PM

Originally Posted by circuscircus

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

Krizalid is saying you could notice that $( \tan x )^{ \sec x} = e^{\sec x \ln ( \tan x)}$ . and find the derivative of that. $\left( \frac d{dx}e^u = u'e^u \right)$

**circuscircus** · Sep 25th 2007, 10:40 PM

$e^{\sec x \ln ( \tan x)} secxtanx * \frac{1}{tan x} * sec^2u<br />$

so like this?

**Jhevon** · Sep 25th 2007, 10:43 PM

Originally Posted by circuscircus

$e^{\sec x \ln ( \tan x)} secxtanx * \frac{1}{tan x} * sec^2u<br />$

so like this?

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

**circuscircus** · Sep 25th 2007, 11:07 PM

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

so would be like this?

**Jhevon** · Sep 25th 2007, 11:27 PM

Originally Posted by circuscircus

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

so would be like this?

use parentheses! you have the idea, but the answer as written is wrong.

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

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

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

LinkBack

Thread Tools

Display

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

Similar Math Help Forum Discussions

derivative of secx

The derivative of (secx)^2?

Simplifying sin(x+4)secx?

integrate (secx)^4

differentia secx????????

Search Tags