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Math Help - Derivative...

  1. #1
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    Derivative...

    Original: If f(x)=2 (Root(x))ln(x), find f'(x).

    u' = 1/2*2 (x^(1/2-1) = x^-1/2= 1/x^1/2
    v' ln(x) = 1/x

    Did the product rule.... and got

    My answer was 1/x^1/2*(ln(x)) + 1/x*2(x^1/2)

    I typically get these types of problems right but I just cant figure out what I did wrong... If anyone can spot my error help would be greatly appreciated!!
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  2. #2
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    Quote Originally Posted by sitdownson View Post
    Original: If f(x)=2 (Root(x))ln(x), find f'(x).

    u' = 1/2*2 (x^(1/2-1) = x^-1/2= 1/x^1/2
    v' ln(x) = 1/x

    Did the product rule.... and got

    My answer was 1/x^1/2*(ln(x)) + 1/x*2(x^1/2)

    I typically get these types of problems right but I just cant figure out what I did wrong... If anyone can spot my error help would be greatly appreciated!!
    f(x)=2\sqrt{x}\ lnx=2uv

    \displaystyle\ u=\sqrt{x}\Rightarrow\ u'=\frac{1}{2\sqrt{x}}

    \displaystyle\ v=lnx\Rightarrow\ v'=\frac{1}{x}

    \displaystyle\ 2\frac{d}{dx}(uv)=2\left(vu'+uv'\right)=\frac{lnx}  {\sqrt{x}}+\frac{2\sqrt{x}}{x}

    which is what you got.
    It can be be simplified a little to

    \displaystyle\ \frac{lnx}{\sqrt{x}}+\frac{2\sqrt{x}}{\sqrt{x}\sqr  t{x}}=\frac{lnx+2}{\sqrt{x}}
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