What is the second derivative of #f(x)=x^2/(x^2+3) #?

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Answer 1 · 2016-01-17T12:31:37

#[18(1-x^2)]/(x^2+3)^3#

Using the quotient rule which states that:

If #f(x) = (u(x))/(v(x))# then #f'(x) = (vu' - uv')/v^2#

#u = x^2, u' = 2x#
#v=x^2+3, v'=2x#

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

#=(6x)/(x^2+3)^2#

Reset u & v.

#u=6x, u'=6#
#v=(x^2+3)^2, v'=4x(x^2+3)#

#f''(x)=((x^2+3)^2*6-6x*4x(x^2+3))/(x^2+3)^4#

#=((x^2+3)[6x^2+18-24x^2])/(x^2+3)^4=[18(1-x^2)]/(x^2+3)^3#