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How to derive the derivative formula of arctan x?

  1. Mar 29, 2007 #1


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    1. The problem statement, all variables and given/known data

    basically what the topic states - derive the formula for the derivative of arctan x.

    2. Relevant equations

    d/dx (arctan x) = 1/(1+x^2)

    3. The attempt at a solution

    strange question because we already know the answer. but im not sure how to start this.

    i know arctan x = y

    therefore tan y = x

    but what can i do with this? do i need to draw a right triangle and label all the sides? can someone help me get started? thanks.
  2. jcsd
  3. Mar 29, 2007 #2
    Do you know the Inverse function theorem? Using it and a right triangle (as you said), gives the result pretty quickly.
  4. Mar 29, 2007 #3
    You could indeed draw a triangle knowing that the tangent is equal to x. In a triangle, if you're given that the tangent of some angle y is equal to x, what do you know about the lengths of the sides?
  5. Mar 30, 2007 #4


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    the opposite would be x, the adjacent is 1 and the hypoteneuse is the square root of 1+x^2

    after that im stuck.
  6. Mar 30, 2007 #5


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    You got tan y = x, you want dy/dx, so use implicit differentiation.
  7. Mar 30, 2007 #6

    Gib Z

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    [tex]y=arctan x[/tex]

    Therefore [tex]x=\tan y[/tex], Quite easy to see.

    [tex]\frac{dx}{dy}=\sec^2 y[/tex]

    Using the Pythagorean Identity [itex]\sec^2 y = \tan^2 y +1[/itex] we can get this: [tex]\frac{dx}{dy}=\tan^2 y +1[/tex].

    Flip the fraction since we want dy/dx, not dx/dy. And also, as seen on my second line tan y=x, so tan^2 y = x^2.

    Thats how we get

  8. Mar 30, 2007 #7


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    [tex] y=\arctan x [/tex]

    [tex] \tan y =x [/tex]

    [tex] \frac{d}{dx}\tan y= \frac{1}{\cos^{2}y}\frac{dy}{dx}=1 [/tex]

    [tex] \frac{dy}{dx} =\cos^{2} y=\left(\frac{1}{\sqrt{1+x^{2}}}\right)^{2}=\frac{1}{1+x^{2}} [/tex]
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