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Arctan derivatives problem

  1. Feb 11, 2006 #1
    Find the derivatives of the following arctan {[(1+x^2)^1/2 ] - 1} / x with respect to arctan x.may i know how to do this??
    how to start and the steps??
     
  2. jcsd
  3. Feb 11, 2006 #2

    arildno

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    Let your independent variable be given as [itex]u=arctan(x)[/itex] that is,
    [itex]x(u)=tan(u)[/itex]

    You have been given the function of x,
    [tex]F(x)=\frac{arctan(\sqrt{1+x^{2}}-1)}{x}[/tex]
    Let [itex]h(u)=F(x(u))[/tex]
    You are asked to find [itex]\frac{dh}{du}[/itex]


    Note that it is easy to give your final answer in terms of "x" rather than "u", since we have [tex]\frac{dx}{du}=\frac{1}{\cos^{2}u}=1+\tan^{2}u=1+x^{2}[/tex]
     
    Last edited: Feb 11, 2006
  4. Feb 11, 2006 #3

    HallsofIvy

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    You have [itex]\frac{arctan(u)}{x}[/itex]. Since that is a quotient of two functions use the "quotient" rule:
    [tex]\frac{d\left(\frac{f(u)}{x}}= \frac{\frac{df}{du}\frac{du}{dx}x- f(u)}{x}[/tex]
    where [itex]u= \sqrt{1+x^2}-1[/itex] and f(u)= arctan u.
    The [itex]\frac{df}{du}\frac{du}{dx} in that is the "chain" rule.
     
  5. Feb 11, 2006 #4
    sotty,for threat 2 i don't really understand.can u plsexplain more clearly??
     
  6. Feb 11, 2006 #5

    arildno

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    HallsofIvy's "u" is not the same as my "u".
    He is presenting a technique to evaluate the term [itex]\frac{dF}{dx}[/itex] in my terms.

    We have: [itex]\frac{d}{dx}(\sqrt{1+x^{2}}-1)=\frac{x}{\sqrt{1+x^{2}}}[/itex]
    Thus, we get:
    [tex]\frac{dF}{dx}=\frac{\sqrt{1+x^{2}}}{1+(\sqrt{1+x^{2}}-1)^{2}}-\frac{F(x)}{x}[/tex]
    Multiply this with [itex]1+x^{2}[/itex] to get [itex]\frac{dh}{du}[/itex]
     
  7. Feb 11, 2006 #6
    can u pls show me the whole steps ??thanx
     
  8. Feb 11, 2006 #7

    arildno

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    Darn, I made a mistake!
    the correct expression for the derivative with respect to u=arctan(x), should be:
    [tex]\frac{dh}{du}=\frac{\sqrt{1+x^{2}}}{1+(\sqrt{1+x^{2}}-1)^{2}}-F(x)\frac{1+x^{2}}{x}[/tex]
     
  9. Feb 11, 2006 #8
    [itex]\\int_1^4 \\sqrt{t}(ln(t))dt[/itex]
     
  10. Feb 11, 2006 #9

    VietDao29

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    Unfortunately, I don't think we are going to provide you a step by step solution. However, we may help you.
    Just tell us where in the post that you don't really understand and we may clarify it for you.
     
  11. Feb 12, 2006 #10

    arildno

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    It might be helpful to remember the following rule:

    Suppose you are to differentiate a function F(x) with respect to a (invertible) function f(x).
    Then we have, in general:
    [tex]\frac{dF}{df}=\frac{\frac{dF}{dx}}{\frac{df}{dx}}[/tex]
     
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