hey maths forum!

how do you differentiate the following:

$\displaystyle y = tan^-1(\frac{1+tanx}{1-tanx}) $

thanks

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- Jan 20th 2007, 11:07 AMdadondifferentiation
hey maths forum!

how do you differentiate the following:

$\displaystyle y = tan^-1(\frac{1+tanx}{1-tanx}) $

thanks - Jan 20th 2007, 12:12 PMgalactus
The derivative of arctan is $\displaystyle \frac{1}{x^{2}+1}$

Using the chain rule:

$\displaystyle \frac{1}{(\frac{1+tan(x)}{1-tan(x)})^{2}+1}\cdot\frac{d}{dx}\left(\frac{1+tan( x)}{1-tan(x)}\right)$

$\displaystyle \frac{1}{(\frac{cos(x)+sin(x)}{cos(x)-sin(x)})^{2}+1}\cdot\frac{2}{(cos(x)-sin(x))^{2}}=1$ - Jan 20th 2007, 12:39 PMticbol
----d/dx arctan(u) = 1 /(1 +u^2) *du/dx

----d/dx tan(u) = sec^2(u) *du/dx

y = arctan[(1 +tanX) /(1 -tanX)]

Differentiate both sides with respect to X,

dy/dX = (1 / [1 +{(1+tanX)/(1-tanX)}^2]) *({[(1-tanX)(sec^2(X)] -[(1+tanX)(-sec^2(X)]} / { (1-tanX)^2})

dy/dX = (1 / {[(1-tanX)^2 +(1+tanX)^2] /(1-tanX)^2}) *({[(1-tanX)(sec^2(X)] +[(1+tanX)(sec^2(X)]} / { (1-tanX)^2})

If you can follow through the thousand brackets/parentheses, you're good!

If I got lost in doing those more than thousand enclosures, then I'm normal.

dy/dX = ([(1-tanX)^2] / [(1-tanX)^2 +(1+tanX)^2]) *({[sec^2(X)]*[(1-tanX) +(1+tanX)]} / { (1-tanX)^2})

dy/dX = ([ 1/ [(1-tanX)^2 +(1+tanX)^2]) *([sec^2(X)]*[2] )

dy/dX = [2sec^2(X)] / [(1 -2tanX +tan^2(X)) +(1 +2tanX +tan^2(X))]

dy/dX = [2sec^2(X)] / [2 +2tan^2(X)]

dy/dX = [sec^2(X)] / [1 +tan^2(X)]

dy/dX = sec^2(X) / sec^2(X)

dy/dX = 1 --------------------------answer.

Heck, after all those million enclosures? - Jan 20th 2007, 12:42 PMdadon
lol !

Sorry I didn't know it was going to be that long. Thanks guys really appreciate it! :)

im going to look over it now... - Jan 20th 2007, 01:19 PMdadon
hey I tried taking that down on paper, and its all over the place. lol

can you recommend a book ? that will be useful for this and general maths at this level.

sorry im useless at maths and need a complete idiot's guide lol :o - Jan 20th 2007, 02:08 PMThePerfectHacker
- Jan 21st 2007, 02:00 AMdadon
hi

ticbol do you use the chain rule? - Jan 21st 2007, 02:52 AMCaptainBlack
In which case cost should be considered. Morris Kline's book is in reprint

by Dover and available at ~$20. See here.

RonL - Jan 21st 2007, 03:15 AMticbol
I do, sometimes, whatever is that chain rule.

[No kidding, I have to look up that, chain rule, from Google. I am not good at definitions/titles/terms/whatever. Like the cummulative, transitive, confusative properties of addition(?) or multiplication(?). I just do the operations/procedures/ways/whatever as I know how to do them. That's why I hate Proofs in Geometry!]

[Oh, that chain rule! Why, of cousre. I have to continue differentiating until the innermost expression with variables is done.

d/dx (3x^2 -2)^3

= 3(3x^2 -2)^2 and then differentiate the inner 3x^2 -2,

= 3(3x^2 -2)*(3*2*x)

That chain rule.]

Or do you mean did I use the chain rule in my solution above?

Umm, I don't know.

As I have said in the beginning, as a reminder,

d/dx arctan(u) = 1/(1+u^2) *du/dx

See that *du/dx at the end? That, I think, follows the chain rule.

Like I said, I am not into definitions/rules/whatnot. I just do what I know is right. And if I am asked why I did "that", then I can always explain it. That means I know the meaning of the definitions/rules/whatnot, but I do not always know how they are worded/standardized/etched-in-stones.

Like most Math guys, I hate memorization. If I have to memorize something, I have to fight with myself. - Jan 21st 2007, 06:26 AMThePerfectHacker
- Jan 21st 2007, 06:52 AMgalactus
I agree, PH. From what I can gather, they are reprints of old textbooks.

- Jan 21st 2007, 08:03 AMCaptainBlack
Usually they are reprints of classic monographs rather than text books.

In this case it is a text book and other than using US customary units

(yuk, yuk ....) is very good.

Kline was a great expositor, even now I can't read his writings without

hearing his very characteristic voice speaking them

RonL - Jan 21st 2007, 09:40 AMSoroban
Hello, dadon!

Seeing that the answer came out to a*constant*, I explored a bit . . .

Quote:

Differentiate: .$\displaystyle y \;= \;\tan^-1\left(\frac{1 +\tan x}{1-\tan x}\right) $

I finally saw that: .$\displaystyle \frac{1 + \tan x}{1 - \tan x} \:=\:\frac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4}\tan x} \:=\:\tan\left(\frac{\pi}{4} + x\right)$

So the function is: .$\displaystyle y \;=\;\tan^{-1}\!\!\left[\tan\left(\frac{\pi}{4} + x\right)\right] \;=\;\frac{\pi}{4} + x + \pi n$

. . Therefore: .$\displaystyle y'\:=\:1$