Letf(x)=x/1+x^2. (a) Find f′(x), simplify your answer as much as possible. (b) Find f′′(x), simplify your answer as much as possible.

Answer 1

You left out parentheses but I'm guessing this is:

If so, use the quotient rule to find f '(x). As a reminder, if f(x) = u/v then f '(x) = (u'v - uv')/v²

u = x

u' = 1

v = 1 + x²

v' = 2x

v² = (1 + x²)²

So:

f '(x) = [(1)(1 + x²) - (x)(2x)]/(1 + x²)²

f '(x) = [1 + x² - 2x²)]/(1 + x²)²

f '(x) = (1 - x²)/(1 + x²)²

For f ''(x), again use the quotient rule. This time:

u = 1 - x²

u' = -2x

v = (1 + x²)²

v' = 2(1 + x²)(2x) = 4x(1 + x²) = 4x + 4x³

v² = (1 + x²)⁴

So:

f ''(x) = [(-2x)(1 + x²)² - (1 - x²)(4x + 4x³)]/(1 + x²)⁴

f ''(x) = [(-2x)(1 + 2x² + x⁴) - (4x + 4x³ - 4x³ - 4x⁵)]/(1 + x²)⁴

f ''(x) = [-2x - 4x³ - 2x⁵ - (4x - 4x⁵)]/(1 + x²)⁴

f ''(x) = (-2x - 4x³ - 2x⁵ - 4x + 4x⁵)/(1 + x²)⁴

f ''(x) = (2x⁵ - 4x³ - 6x)/(1 + x²)⁴

f ''(x) = 2x(x⁴ - 2x² - 3)/(1 + x²)⁴

f ''(x) = 2x(x² - 3)(x² + 1)/(1 + x²)⁴

f ''(x) = 2x(x² - 3)/(1 + x²)³