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Math Help - [SOLVED] Limiting process for differentiation

  1. #1
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    [SOLVED] Limiting process for differentiation

    From first principles (limiting process), find the derivative of ;
    y= 2/( x^4 +1) + 3/ x

    I would greatly appreciate this help... I tried to work it out but it's confusing and long. Please show details.
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  2. #2
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    Hello, thaliaj_df!

    This is a messy problem . . .
    They could have easily made two problems out of it.


    From first principles, find the derivative of: . y \:=\:\frac{2}{x^4 +1} + \frac{3}{x}
    I must assume that you know the four steps of the First Principles:

    . . (1) Find f(x+h) . . . Replace x with x+h ... and simplify.

    . . (2) Subtract  f(x) . . . Subtract the original function ... and simplify.

    . . (3) Divide by h . . . Factor and reduce.

    . . (4) Take the limit as h\to0 . . . Let h = 0 and see "what's left."


    We will differentiate the two fractions separately . . .


    Let f(x) \:=\:\frac{2}{x^4+1}

    (1)\;\;f(x+h) \:=\:\frac{2}{(x+h)^4 + 1}

    (2)\;\;f(x+h) - f(x) \:=\:\frac{2}{(x+h)^4 + 1} +\frac{2}{x^4+1} \;=\;2\cdot\frac{x^4+1) - [(x+h)^4 - 1]}{(x^4+1)[(x+h)^4+1]}

    . . . . . . . . . . . . =\;2\cdot\frac{x^4+1 - x^4 - 4x^3h - 6x^2h^2 - 4xh^3 - h^4 - 1}{(x^4+1)[(x+h)^4+1]}

    . . . . . . . . . . . . = \;2\cdot\frac{-4x^3h - 6x^2h^2 - 4xh^3 - h^4}{(x^4+1)[(x+h)^4+1]}

    . . . . . . . . . . . . = \;-2\cdot\frac{h(4x^3 + 6x^2h + 4xh^2 + h^3)} {(x^4+1)[(x+h)^4+1]} <br />

    (3)\;\;\frac{f(x+h)-f(x)}{h} \;=\;-2\cdot\frac{4x^3 + 6x^2h + 4xh^2 + h^3}{(x^4+1)[(x+h)^4+1]}

    (4)\;\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;\lim_{h\to0}\left[-2\cdot\frac{4x^3 +6x^2h + 4xh^2 + h^3}{(x^4+1)[(x+h)^4+1]}\right]

    . . . . . . . . . . . . . . . = \;-2\cdot\frac{4x^3 + 0 + 0 + 0}{(x^4+1)[(x+0)^4+1]}

    Hence: . f'(x) \;=\;\frac{\text{-}8x^3}{(x^4+1)^2}



    Let g(x) \:=\:\frac{3}{x}

    (1)\;\;g(x+h) \;=\;\frac{3}{x+h}

    (2)\;\;g(x+h) - g(x) \;=\;\frac{3}{x+h} - \frac{3}{x} \;=\;\frac{\text{-}3h}{x(x+h)}

    (3)\;\;\frac{g(x+h)-g(x)}{h} \;=\;\frac{\text{-}3}{x(x+h)}

    (4)\;\;\lim_{h\to0}\frac{\text{-}3}{x(x+h)} \;=\;\frac{\text{-}3}{x(x+0)}

    Hence: . g'(x) \;=\;\frac{\text{-}3}{x^2}


    Therefore: . \frac{dy}{dx} \;=\;f'(x) - g'(x) \;=\;\frac{\text{-}8x^3}{(x^4+1)^2} - \left(\frac{\text{-}3}{x^2}\right) \;=\;\frac{3}{x^2} - \frac{8x^3}{(x^4+1)^2}

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