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Math Help - help with a derivative involving arctan

  1. #1
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    help with a derivative involving arctan

    find derivative of
    f(x)=(sq.rt.)[x] - arctan(sq.rt.)[x]

    my book says ((sq.rt.)(x))/(2(1+x)
    however, online i found it to be (1/(2(sq.rt.)[x])) - (1/(2(sq.rt.)[x](x-1)))

    i tried to simplify either one into the other, but no luck, so that means one of them is wrong...however, i cant find the derivative initially, so it doesnt matter

    i would appreciate it if anybody could help me solve this
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  2. #2
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    Quote Originally Posted by twostep08 View Post
    find derivative of
    f(x)=(sq.rt.)[x] - arctan(sq.rt.)[x]

    my book says ((sq.rt.)(x))/(2(1+x)
    however, online i found it to be (1/(2(sq.rt.)[x])) - (1/(2(sq.rt.)[x](x-1)))

    i tried to simplify either one into the other, but no luck, so that means one of them is wrong...however, i cant find the derivative initially, so it doesnt matter

    i would appreciate it if anybody could help me solve this
    f(x) = \sqrt{x} - \arctan(\sqrt{x})

    f'(x) = \frac{1}{2\sqrt{x}} - \frac{\frac{1}{2\sqrt{x}}}{1 + x}

    f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{x}(1 + x)}

    f'(x) = \frac{1+x}{2\sqrt{x}(1+x)} - \frac{1}{2\sqrt{x}(1 + x)}

    f'(x) = \frac{x}{2\sqrt{x}(1+x)} = \frac{\sqrt{x}}{2(1+x)}
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  3. #3
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    makes sense, however howd you get derivative of arctan (sq.rt.[x]) to be
    <br />
    " alt="\frac{\frac{1}{2\sqrt{x}}}{1 + x}
    " />
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  4. #4
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    Quote Originally Posted by twostep08 View Post
    makes sense, however howd you get derivative of arctan (sq.rt.[x]) to be
    <br />
    " alt="\frac{\frac{1}{2\sqrt{x}}}{1 + x}
    " />
    \frac{d}{dx} \arctan(u) = \frac{u'}{1+u^2}<br />

    in this case, u = \sqrt{x}
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