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Thread: [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: .$\displaystyle y \:=\:\frac{2}{x^4 +1} + \frac{3}{x}$
    I must assume that you know the four steps of the First Principles:

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

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

    . . (3) Divide by $\displaystyle h$ . . . Factor and reduce.

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


    We will differentiate the two fractions separately . . .


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

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

    $\displaystyle (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]} $

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

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

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

    $\displaystyle (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]} $

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

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

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



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

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

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

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

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

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


    Therefore: .$\displaystyle \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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