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

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Answer 1 · 2018-05-25T13:24:39

Below

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

Answer 2 · 2018-05-25T13:28:41

Please see the explanation below

#(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#