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Differentiating arctan

  1. Apr 2, 2004 #1
    this thing is ugly

    I was given this problem yesterday. It's asking me to differentiate a function containing arctan. Should I find a common denominator and combine them into one term, should I just differentiate each term separately? Please I need help with this thing! I've been told it comes out to 0 but I'm not sure if he is correct or not. Thanks for replying if you choose to!
     
  2. jcsd
  3. Apr 2, 2004 #2

    Doc Al

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    Should you try to combine the two arctan terms into one? Yikes! :eek:

    No, just differentiate each term separately. If you know how to differentiate arctan(x) and how to use the chain rule, this should be a snap.
     
  4. Apr 2, 2004 #3

    ShawnD

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    According to maple, f'(x) is this



    [tex]\left( 1+{x}^{2} \right) ^{-1}-{\frac {1}{{x}^{2} \left( 1+{x}^{-2} \right) }}[/tex]

    I can't help with how to get that though; I've never dealt with arctan before.

    It simplifies to 0.
     
    Last edited: Apr 2, 2004
  5. Apr 2, 2004 #4

    chroot

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  6. Apr 2, 2004 #5

    Janitor

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    Draw a right triangle with a base of length 1 and a side of length x. Note that one of the acute angles is equal to arctan(x) and the other acute angle is equal to arctan(1/x). Since the three interior angles of any triangle add up to pi radians, and since the remaining angle in this triangle is pi/2, then the sum arctan(x) + arctan(1/x) has to equal the constant value pi - pi/2 = pi/2. Its derivative is thus zero.
     
  7. Apr 2, 2004 #6
    [tex]y = \arctan{x}[/tex]
    [tex]\tan{y} = x[/tex]
    [tex]\frac{dy}{dx}\sec^2{y} = 1[/tex]
    [tex]\frac{dy}{dx} = \cos^2{y}[/tex]

    Draw a triangle based on the second line, tan(y) = x. This means that the opposite side is x and the adjacent side is 1, giving the hypotenuse a value of [itex]\sqrt{x^2 + 1}[/itex]. This means that cos(y) will be 1 over the hypotenuse, i.e.

    [tex]\cos{y} = \frac{1}{\sqrt{x^2 + 1}}[/tex]

    which means that our derivative is

    [tex]\frac{dy}{dx} = \frac{1}{x^2 + 1}[/tex]
    [tex]\frac{d}{dx}\Big(\arctan{x}\Big) = \frac{1}{x^2 + 1}[/tex]

    The rest follows from the chain rule.

    cookiemonster
     
  8. Apr 2, 2004 #7
    ok i think I can see what you have done. can you explain how you got from tan(y)=x to the line below it? did you just differentiate both sides with respect to y?

    And then at the end are you saying to just substitute the equation for the derivative of arctan into the orginal equation?

    Thanks for all your help guys!
     
    Last edited: Apr 2, 2004
  9. Apr 2, 2004 #8
    Yes. The third line is reached by differentiating the second with respect to x.

    The derivative of arctan(1/x) is found in a similar manner, or you can differentiate it using a substitution and the chain rule. Both methods require just about the same amount of work.

    cookiemonster
     
  10. Apr 2, 2004 #9

    Janitor

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    I had a buddy in high school who liked to use non-calculus techniques to get the answer to homework problems that the calculus teacher assigned. The kid knew darn well that the teacher would not give credit for getting the right answer by "devious" means, but that didn't stop him from doing that. I think he had some sort of persecution complex, and he loved to wail to anyone who would listen about how unfair the teacher was.
     
  11. Apr 2, 2004 #10
    Heh, just read your solution, Janitor. That's pretty neat.

    cookiemonster
     
  12. Apr 2, 2004 #11

    Janitor

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    Thanks, Cookiemonster! But I had a guilty conscience since I figured the teacher wouldn't give him any credit for it, since it doesn't teach him how to do derivatives.
     
  13. Apr 2, 2004 #12
    Haha, why not? He still has to differentiate pi/2. =]

    cookiemonster
     
  14. Apr 2, 2004 #13

    Janitor

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    True!

    . .
     
  15. Apr 3, 2004 #14
    ok so the derivative of arctan x is 1/[(x^2)+1] but what is the derivative of arctan(1/x) do i just plug 1/x into 1/[(x^2)+1]? it seems like the logical thing to do but then again, i may be totally wrong.. Thanks.
     
  16. Apr 3, 2004 #15
    As I said in my previous post, you can figure it out using the same method as for arctan(x), i.e.

    tan(y) = 1/x

    differentiate both sides implicitly to get

    dy/dx = -cos^2(y)/x^2

    then we build a triangle based on tan(y) = 1/x, which means the opposite side is 1 and the adjacent side is x, so the hypotenuse is sqrt(x^2 + 1), so the cos^2(y) term is x/sqrt(x^2 + 1), meaning

    dy/dx = 1/x^2 * -x^2/(1+x^2) = -1/(x^2 + 1)

    Alternatively, you could have made a substitution u = 1/x into the derivative, so we have

    d/dx(arctan(u)) = df(u)/du * du/dx = 1/(1 + 1/x^2)*-1/x^2

    which can be shown to be equivalent with a little bit of algebra.

    cookiemonster
     
  17. Apr 3, 2004 #16
    so the derivative of the problem simplifies to -1/(x^2 + 1)? and that equals 0?
     
  18. Apr 3, 2004 #17
    Yes. Both Janitor's solution and brute force show the solution to be 0. You wanted it to be 0 in the first place anyway, didn't you?

    cookiemonster
     
  19. Apr 4, 2004 #18
    alright, i understand everything now except this one line:
    dy/dx = 1/x^2 * -x^2/(1+x^2) = -1/(x^2 + 1)

    how did you get this? i got up to dy/dx = -x^2+1/x^2
    whatd you do to get that line? i tried using the quotient rule to differentiate it but i got 2x/x^4 but that is incorrect.

    i never wanted the answer to be zero, i was just told that it is. i guess. thanks alot!
     
    Last edited: Apr 4, 2004
  20. Apr 4, 2004 #19

    Doc Al

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    That's the derivative of the second term in your original expression. Add that to the derivative of the first term and the sum will equal zero.
     
  21. Apr 4, 2004 #20

    Doc Al

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    I assume you can differentiate arctan(x)? That will be the first term of your answer.

    Now use the chain rule to differentiate arctan(u), where u = 1/x. (What's the derivative of 1/x?) That will be the second term making up your answer.
     
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