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Math Help - differentiation

  1. #1
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    differentiation

    hey maths forum!


    how do you differentiate the following:


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


    thanks
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  2. #2
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    The derivative of arctan is \frac{1}{x^{2}+1}

    Using the chain rule:

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

    \frac{1}{(\frac{cos(x)+sin(x)}{cos(x)-sin(x)})^{2}+1}\cdot\frac{2}{(cos(x)-sin(x))^{2}}=1
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  3. #3
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    Quote Originally Posted by dadon View Post
    hey maths forum!


    how do you differentiate the following:


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


    thanks
    ----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?
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  4. #4
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    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...
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  5. #5
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    Question

    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
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  6. #6
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    Quote Originally Posted by dadon View Post
    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
    One thing I realized that most Calculus books are the same. Just buy any.
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  7. #7
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    hi

    ticbol do you use the chain rule?
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    One thing I realized that most Calculus books are the same. Just buy any.
    In which case cost should be considered. Morris Kline's book is in reprint
    by Dover and available at ~$20. See here.

    RonL
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  9. #9
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    Quote Originally Posted by dadon View Post
    hi

    ticbol do you use the chain rule?
    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.
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  10. #10
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    Quote Originally Posted by CaptainBlank View Post
    In which case cost should be considered. Morris Kline's book is in reprint
    by Dover and available at ~$20. See here.

    RonL
    I never read the Book thus I cannot say if it was good or bad. But I never liked Dover books. They are far too complicated, they already assume you have knowledge of the material before (at least the ones I brought).
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  11. #11
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    I agree, PH. From what I can gather, they are reprints of old textbooks.
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  12. #12
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    Quote Originally Posted by ThePerfectHacker View Post
    I never read the Book thus I cannot say if it was good or bad. But I never liked Dover books. They are far too complicated, they already assume you have knowledge of the material before (at least the ones I brought).
    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
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  13. #13
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    Hello, dadon!

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


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

    I finally saw that: . \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: . y \;=\;\tan^{-1}\!\!\left[\tan\left(\frac{\pi}{4} + x\right)\right] \;=\;\frac{\pi}{4} + x + \pi n


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

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