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how do you differentiate the following:
$\displaystyle y = tan^-1(\frac{1+tanx}{1-tanx}) $
thanks
The derivative of arctan is $\displaystyle \frac{1}{x^{2}+1}$
Using the chain rule:
$\displaystyle \frac{1}{(\frac{1+tan(x)}{1-tan(x)})^{2}+1}\cdot\frac{d}{dx}\left(\frac{1+tan( x)}{1-tan(x)}\right)$
$\displaystyle \frac{1}{(\frac{cos(x)+sin(x)}{cos(x)-sin(x)})^{2}+1}\cdot\frac{2}{(cos(x)-sin(x))^{2}}=1$
----d/dx arctan(u) = 1 /(1 +u^2) *du/dxOriginally Posted by dadon
----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?
One thing I realized that most Calculus books are the same. Just buy any.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
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.
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
Hello, dadon!
Seeing that the answer came out to a constant, I explored a bit . . .
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: .$\displaystyle y'\:=\:1$
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