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Math Help - Chain rule question #3

  1. #1
    Junior Member madmax29's Avatar
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    Chain rule question #3

    if you have

    y=SQUAREROOT (x^2 + 3)

    let z = (x^2 + 3)

    ...1 dy/dz= 1/2 z ^-1/2 and dz/dx= 2 x <<< can someone explain where the 2 is derived from?

    ...2 then dy/dz x dz/dx = 1/2 z ^-1/2 (2 x)

    ...3 this = x/squareroot z = x/SQUAREROOT (x^2 + 3)

    how do you get from line 2 to line 3 as x/SQUAREROOT (x^2 + 3)
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  2. #2
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    e^(i*pi)'s Avatar
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    Line 1: 2x comes from taking the derivative of x^2+3 which the chain rule requires.

    From line 2 you have \dfrac{dy}{dx} = \dfrac{1}{2} (x^2+3)^{-1/2}

    From the laws of exponents and surds we know that a^{-1/n} = \dfrac{1}{\sqrt[n]{a}}

    Thus \dfrac{1}{2} (x^2+3)^{-1/2} = \dfrac{1}{2} \cdot \dfrac{1}{\sqrt{x^2+3}}

    When multiplying by 2x the 1/2 will cancel leaving you with \dfrac{x}{\sqrt{x^2+3}}

    Which is line 3
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