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B Really an easy question about derivative of arctan

  1. Apr 13, 2017 #1
    Last edited: Apr 13, 2017
  2. jcsd
  3. Apr 13, 2017 #2

    Math_QED

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  4. Apr 13, 2017 #3
    whoops sorry somehow I put it in sqrt.

    Can you explain me why we are considering one specific trigonometry although we could consider a triangle of which hypotenuse is x and base is 1


    Ah please don't mind. I figured out.
     
    Last edited: Apr 13, 2017
  5. Apr 13, 2017 #4

    Mark44

    Staff: Mentor

    I'm not sure what it is that you figured out, but here's the answer to your first question, above.
    Let ##y = \arctan(x)##
    An equivalent equation is ##\tan(y) = x##
    The picture below shows a right triangle with one angle labeled y. Since ##\tan(y) = x##, it's reasonable to label the opposite side as x, and the adjacent side as 1. This forces the hypotenuse to be ##\sqrt{x^2 + 1}##.

    If we differentiate both sides of the equation ##\tan(y) = x## with respect to x, we get ##\sec^2(y) \frac{dy}{dx} = 1##, or ## \frac{dy}{dx} = \frac 1{\sec^2(y)} = \cos^2(y)##. From the drawing, it can be seen that ##\cos^2(y) = \frac 1 {x^2 + 1}##
    triangle.png
     
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