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Thread: Derivative of a trig function

  1. #1
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    Derivative of a trig function



    I'm not sure how to solve this type of problem. Any help would be appreciated. Thanks.
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  2. #2
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    Quote Originally Posted by b521 View Post


    I'm not sure how to solve this type of problem. Any help would be appreciated. Thanks.
    You just need to apply the chain rule. The general formula for

    \frac{d}{dx}Arctan(u)=\frac{u'}{1+u^2}

    Where u is a function of x. You're basically applying the chain rule here. Just remember that the derivative of u=\sqrt{x} is u'=\frac{1}{2\sqrt{x}}, and u^2=x. Once you put it all together, you should obtain:

    \frac{1}{2\sqrt{x}(1+x)}
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  3. #3
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    Thanks! It makes much more sense to me now!
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