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Thread: derivative question.

  1. #1
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    derivative question.

    if f in sqrt(f(x)) is differentiable, would it equal to

    f'(x)
    ------
    2sqrt[f(x)] ?
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  2. #2
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    trying this out with x^3, i get 2(x^3)-1*(3x^2), and i think this would yield....

    3x^2
    ------
    2(x^3)

    I know this is a dumb simplification problem but i just wanted some clarification. thanks.
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Evan.Kimia View Post
    if f in sqrt(f(x)) is differentiable, would it equal to

    f'(x)
    ------
    2sqrt[f(x)] ?
    Let $\displaystyle y = \sqrt{f(x)}$

    $\displaystyle y^2 = f(x)$

    $\displaystyle 2y\frac{dy}{dx} = f'(x)$

    $\displaystyle \frac{dy}{dx} = \frac{f'(x)}{2y} = \frac{f'(x)}{2\sqrt{f(x)}}$

    I get the same answer as you but I'm not 100% confident it's right
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  4. #4
    MHF Contributor ebaines's Avatar
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    Quote Originally Posted by Evan.Kimia View Post
    trying this out with x^3, i get 2(x^3)-1*(3x^2), and i think this would yield....

    3x^2
    ------
    2(x^3)

    I know this is a dumb simplification problem but i just wanted some clarification. thanks.
    Not quite right. You should have:

    $\displaystyle
    f(x) = \sqrt {x^3}
    $
    $\displaystyle
    f'(x) = \frac {3x^2} {2 \sqrt{x^3}} = \frac 3 2 \sqrt x
    $

    An easier way is to realize that
    $\displaystyle \sqrt {x^3} = x^{3/2}
    $

    So you get right to:
    $\displaystyle
    f'(x) = \frac 3 2 x^{1/2} = \frac 3 2 \sqrt x
    $
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