How do you find the first and second derivatives of f(x)=(x)/(x^2+1)f(x)=xx2+1 using the quotient rule?

2 Answers
May 25, 2018

Below

Explanation:

f(x)=x/(x^2+1)f(x)=xx2+1

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

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

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

f''(x)=((x^2+1)^2(-2x)-(1-x^2)*(x^2+1)(2x))/(x^2+1)^4

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

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

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

The quotient rule is given by:

f(x)=u/v

f'(x)=(vu'-uv')/v^2

May 25, 2018

Please see the explanation below

Explanation:

The quotient rule is

(u/v)'=(u'v-uv')/(v^2)

Here,

u=x, =>, u'=1

v=x^2+1, =>, v'=2x

Therefore, the first derivative is

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

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

For the second derivative,

u=(1-x^2), =>, u'=-2x

v=(x^2+1)^2, =>, v'=4x(x^2+1)

Therefore, the second derivative is

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

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

=(2x^3-6x)/(x^2+1)^3

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