# Thread: [SOLVED] Limiting process for differentiation

1. ## [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.

2. 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]}
$

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

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

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

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