How do you find the derivative of #h(x) = (x^(1/3)) / (x^2+1)#?

Question

1554 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-09T18:10:55

#1/(3x^(2/3)(x^2 +1)) - (2x^(4/3))/ ((x^2 +1)^2#

With #h(x) = (x^(1/3))/(x^2+1)# you use the quotient rule, which states:

#(f/g)'= (g f' - f g')/ g^2 #

Derivative of #f(x)=x^(1/3)#:

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

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

Derivative of #g(x)=(x^2+1)#:

#g'(x)=2x#

Now plug in:

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

#((x^2+1)/(3x^(2/3)) - 2x^(4/3))/ ((x^2 +1)^2#

#((x^2+1)/(3x^(2/3)))/(x^2 +1)^2 - (2x^(4/3))/ ((x^2 +1)^2#

#1/(3x^(2/3)(x^2 +1)) - (2x^(4/3))/ ((x^2 +1)^2#