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Math Help - [SOLVED] derivative

  1. #1
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    [SOLVED] derivative

    f(x)= 1/sq rt(x) + 2
    find y '

    is this answer -1 and if so if the sq root had x^2 under it would the answer be -2
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  2. #2
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    Which is correct?

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

    2. f(x) = \frac{1}{\sqrt x + 2}
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  3. #3
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    number 2
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  4. #4
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    f(x) = \frac{1}{\sqrt x + 2}

    f'(x) = \frac{(\sqrt x + 2)(0) - (1)(\frac{1}{2\sqrt x})}{(\sqrt x + 2)^2}

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

    f'(x) = -\frac{1}{2(\sqrt x)(\sqrt x + 2)^2}
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  5. #5
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    how did you get the 1/ 2sqrt(x) in the numerator, do you chand the sq rt to x^2
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  6. #6
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    The derivative of a function in the form \frac{f(x)}{g(x)} is \frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{(g(x))^2}. This is known as the quotient formula for derivatives.
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  7. #7
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    yea i was trying to figure out the process step by step on how sq rt (x) equals 2sq rt(x)
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  8. #8
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    The derivative of \sqrt x + 2 is given by the power rule:
    \frac{d}{dx} x^n = nx^{n-1}.

    What you have is x^{1/2} + 2.

    Taking the derivatives:

    \frac{d}{dx} x^{1/2} = (1/2)x^{-1/2} = \frac{1}{2\sqrt x}

    \frac{d}{dx} 2 = 0

    \frac{d}{dx} (\sqrt x + 2) = \frac{1}{2\sqrt x}
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