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Math Help - please help with derivatives of inverse trig functions

  1. #1
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    please help with derivatives of inverse trig functions

    the problem is x arctan squareroot(x)


    i solve it out and got (square root of x)/2(1+x)


    However, the book answer gives it as my answer + arctan(square root of x)


    How did the book get the + arctan(squre root of x)

    I went through my work 4 times and I can't find out why. Unless the x in fron of arctan square root of x plays a role via chain rule.


    If the chain rule is the case, what happens when there is a constant, such as 29 instead of x, so the problem is 29 arctan (squareroot of x).

    Would it thus have a + 29 arctan (square root of x)?
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  2. #2
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    Quote Originally Posted by masterofcheese View Post
    the problem is x arctan squareroot(x)


    i solve it out and got (square root of x)/2(1+x)


    However, the book answer gives it as my answer + arctan(square root of x)


    How did the book get the + arctan(squre root of x)

    I went through my work 4 times and I can't find out why. Unless the x in fron of arctan square root of x plays a role via chain rule.

    \textcolor{red}{y = x\arctan{\sqrt{x}}}

    product rule ...

    \textcolor{red}{y' = x \cdot \frac{1}{2\sqrt{x}(1 + x)} + \arctan{\sqrt{x}} \cdot 1}<br />
    ...
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  3. #3
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    Remember it's a product and (fg)'=f'g+fg' where f(x)=x and g(x)=\arctan (\sqrt{x} )
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