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Math Help - the derivative of arctan

  1. #1
    Member Jones's Avatar
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    the derivative of arctan

    Hi,

    Hm, trying to get my head around this

    the derivative of arctan is \frac{1}{1+x^2}

    So the derivative of arctan(0.5sinx) should be:

    \frac{cosx}{1+\frac{sin^2x}{4}} But that's apparently wrong.

    Why?
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  2. #2
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    Quote Originally Posted by Jones View Post
    Hi,

    Hm, trying to get my head around this

    the derivative of arctan is \frac{1}{1+x^2}

    So the derivative of arctan(0.5sinx) should be:

    \frac{cosx}{1+\frac{sin^2x}{4}} But that's apparently wrong.

    Why?
    Hi, be careful when you're using the chain rule.

     \frac{d}{dx} \arctan \left(\frac{1}{2} \sin (x) \right) = \frac{d \arctan (u)}{du} \times \frac{du}{dx} where  u = 0.5 \sin (x) and \frac{d \arctan (u)}{du} = \frac{1}{1+u^2}

    The denominator of your answer is correct. But look at what you did to the numerator. The numerator is  \frac{du}{dx} = \frac{d}{dx} 0.5 \sin (x) = 0.5 \cos (x) The 0.5 does not dissapear. That is your only mistake.
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    Member Jones's Avatar
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