What is the second derivative of #f(x)=(x-1)/(x^2+1)#?

Question

Luke Alvarez · Accepted Answer

# f''(x)={2(1+x)(1-4x+x^2)}/(1+x^2)^3.#

#f(x)=(x-1)/(x^2+1).#

Using the Quotient Rule for Diffn., we get, the first deri.,

#f'(x)={(x^2+1)d/dx(x-1)-(x-1)d/dx(x^2+1)}/(x^2+1)^2#

#={(x^2+1)(1)-(x-1)(2x)}/(x^2+1)^2={x^2+1-2x^2+2x}/(x^2+1)^2#

#:. f'(x)=(1+2x-x^2)/(1+2x^2+x^4)#

Rediff.ing #f'(x)# to get second deri.,

#f''(x)={(1+x^2)^2d/dx(1+2x-x^2)-(1+2x-x^2)d/dx(1+2x^2+x^4)}/{(x^2+1)^2}^2#

#={(1+x^2)^2(2-2x)-(1+2x-x^2)(4x+4x^3)}/(x^2+1)^4#

#={2(1+x^2)^2(1-x)-4x(1+x^2)(1+2x-x^2)}/(x^2+1)^4#

#=[2(1+x^2){(1+x^2)(1-x)-2x(1+2x-x^2)}]/(x^2+1)^4#

#=[2(1+x^2){1+x^2-x-x^3-2x-4x^2+2x^3}]/(x^2+1)^4#

#:. f''(x)={2(1-3x-3x^2+x^3)}/(1+x^2)^3.#

#=[2{(1+x^3)-3x(1+x)}]/(1+x^2)^3#

#=[2{(1+x)(1-x+x^2)-3x(1+x)}]/(1+x^2)^3#

#=[2(1+x)(1-x+x^2-3x)}/(1+x^2)^3#

#rArr f''(x)={2(1+x)(1-4x+x^2)}/(1+x^2)^3#

Enjoy Maths.!

Emily Ammerman · Answer

undefined To find the second derivative of $ f(x) = \frac{x - 1}{x^2 + 1} $, you first need to find the first derivative, then differentiate it again using the rules of differentiation.

First, find the first derivative:

$$ f'(x) = \frac{d}{dx}\left(\frac{x - 1}{x^2 + 1}\right) $$

Using the quotient rule:

$$ f'(x) = \frac{(x^2 + 1)\frac{d}{dx}(x - 1) - (x - 1)\frac{d}{dx}(x^2 + 1)}{(x^2 + 1)^2} $$

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

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

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

Now, differentiate $ f'(x) $ again to find the second derivative:

$$ f''(x) = \frac{d}{dx}\left(\frac{2x^2 + 2x + 1}{(x^2 + 1)^2}\right) $$

Using the quotient rule:

$$ f''(x) = \frac{(x^2 + 1)^2 \cdot \frac{d}{dx}(2x^2 + 2x + 1) - (2x^2 + 2x + 1) \cdot \frac{d}{dx}(x^2 + 1)^2}{(x^2 + 1)^4} $$

$$ f''(x) = \frac{(x^2 + 1)^2 \cdot (4x + 2) - (2x^2 + 2x + 1) \cdot 2(x^2 + 1) \cdot 2x}{(x^2 + 1)^4} $$

$$ f''(x) = \frac{(4x^3 + 4x^2 + 4x + 2x^2 + 2 - 4x^3 - 4x^2 - 8x)}{(x^2 + 1)^3} $$

$$ f''(x) = \frac{6x^2 - 6x + 2}{(x^2 + 1)^3} $$

So, the second derivative of $ f(x) = \frac{x - 1}{x^2 + 1} $ is $ f''(x) = \frac{6x^2 - 6x + 2}{(x^2 + 1)^3} $.

Evelyn Agard · Answer

undefined To find the second derivative of the function $ f(x) = \frac{x - 1}{x^2 + 1} $, we first need to find the first derivative and then differentiate it again.

1. Find the first derivative $ f'(x) $:
$$ f'(x) = \frac{d}{dx}\left(\frac{x - 1}{x^2 + 1}\right) $$

Using the quotient rule:
$$ f'(x) = \frac{(x^2 + 1)(1) - (x - 1)(2x)}{(x^2 + 1)^2} $$

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

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

2. Now, find the second derivative $ f''(x) $:
$$ f''(x) = \frac{d}{dx}\left(\frac{-x^2 + 2x + 1}{(x^2 + 1)^2}\right) $$

Using the quotient rule again:
$$ f''(x) = \frac{(x^2 + 1)^2(-2x + 2) - (-x^2 + 2x + 1)(2x + 2x)}{(x^2 + 1)^4} $$

$$ f''(x) = \frac{(x^4 + 2x^2 + 1)(-2x + 2) - (-2x^3 - 2x^2 + 2x^3 + 4x^2 - 2x)}{(x^2 + 1)^4} $$

$$ f''(x) = \frac{(-2x^5 + 2x^4 - 4x^3 + 4x^3 + 8x^2 + 2x^2 - 2x + 2)}{(x^2 + 1)^4} $$

$$ f''(x) = \frac{-2x^5 + 2x^4 + 10x^2 - 2x + 2}{(x^2 + 1)^4} $$

So, the second derivative of $ f(x) = \frac{x - 1}{x^2 + 1} $ is $ f''(x) = \frac{-2x^5 + 2x^4 + 10x^2 - 2x + 2}{(x^2 + 1)^4} $.