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Thread: 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 $\displaystyle \frac{1}{1+x^2}$

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

    $\displaystyle \frac{cosx}{1+\frac{sin^2x}{4}}$ 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 $\displaystyle \frac{1}{1+x^2}$

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

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

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

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