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Find the derivative of f(x) = (square root of x)(ln x).

Using my notes, I came up with f'(x) = (1/2*square root of x)/x

The answer in the back of the book says f'(x) = (2+ln x)/(2*square root of x).

Can someone please explain how? Thanks.

  • Calculus/Logarithms - ,

    I think I figured it out. I believe it's just a matter of applying the Chain Rule.

  • Calculus/Logarithms - ,

    y = (x^.5) ln x
    dy/dx = (x^.5) /x + ln x (.5 x^-.5)
    but x^.5 / x = x^-.5 so
    = (x^-.5) (1+.5 ln x)
    = (1+.5 ln x) / x^.5
    = (2+1 ln x)/2 x^.5
    so I agree with the book.

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