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

Question

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

Emily Ammerman · Accepted Answer

#f''(x)=(2x(x^2-3))/(x^2+1)^3#

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

By Quotient Rule,

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

By Quotient Rule,

#f''(x)=(-2x cdot(x^2+1)^2-(1-x^2)cdot2(x^2+1)(2x))/(x^2+1)^4#

By factoring out #2x(x^2+1)# from the numerator,

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

By cancelling out #(x^2+1)#,

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

I hope that this was clear.

Aurora Alvarado · Answer

undefined To find the second derivative of $ f(x) = \frac{x}{x^2 + 1} $, follow these steps:

1. Differentiate $ f(x) $ with respect to $ x $ using the quotient rule to find the first derivative $ f'(x) $.
2. Once you have $ f'(x) $, differentiate it again with respect to $ x $ to find the second derivative $ f''(x) $.

Applying these steps:

1. The first derivative $ f'(x) $ of $ f(x) $ is:
$$ f'(x) = \frac{(x^2 + 1) \cdot 1 - x \cdot 2x}{(x^2 + 1)^2} $$

2. Simplify the expression to get:
$$ f'(x) = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} $$
$$ f'(x) = \frac{1 - x^2}{(x^2 + 1)^2} $$

3. Now, differentiate $ f'(x) $ with respect to $ x $ again to find the second derivative $ f''(x) $. Applying the quotient rule again:
$$ f''(x) = \frac{(2x) \cdot (x^2 + 1)^2 - (1 - x^2) \cdot 2(x^2 + 1)(2x)}{(x^2 + 1)^4} $$

4. Simplify the expression to get the second derivative:
$$ f''(x) = \frac{2x(x^2 + 1)^2 - 2(1 - x^2)(2x)(x^2 + 1)}{(x^2 + 1)^4} $$
$$ f''(x) = \frac{2x(x^2 + 1)^2 - 4x(x^2 + 1)(1 - x^2)}{(x^2 + 1)^4} $$

5. Further simplify the expression to obtain the final form of the second derivative:
$$ f''(x) = \frac{2x(x^2 + 1)^2 - 4x(x^2 + 1) + 4x^3(x^2 + 1)}{(x^2 + 1)^4} $$
$$ f''(x) = \frac{2x(x^4 + 2x^2 + 1) - 4x(x^2 + 1) + 4x^3(x^2 + 1)}{(x^2 + 1)^4} $$
$$ f''(x) = \frac{2x^5 + 4x^3 + 2x - 4x^3 - 4x + 4x^5 + 4x^3}{(x^2 + 1)^4} $$
$$ f''(x) = \frac{6x^5 + 2x}{(x^2 + 1)^4} $$

So, the second derivative of $ f(x) = \frac{x}{x^2 + 1} $ is $ f''(x) = \frac{6x^5 + 2x}{(x^2 + 1)^4} $.

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