Use the quotient rule to find the first derivative #(u/v)' = (vu' - uv')/v^2#
Let #u = (x^2-1)^3, u' = 3(x^2-1)^2 (2x) = 6x(x^2-1)^2#
Let #v = x^2+1, v' = 2x#
#f'(x) = ((x^2+1)*6x(x^2-1)^2 - (x^2-1)^3*2x)/(x^2+1)^2#
Factor: #f(x') = (2x(x^2-1)^2[3(x^2+1) - (x^2-1)])/(x^2+1)^2#
Simplify: #f'(x) = (2x(x^2-1)^2[3x^2+3-x^2+1])/(x^2+1)^2#
Simplify: #f'(x) = (2x(x^2-1)^2(2x^2+4))/(x^2+1)^2#
Simplify: #f'(x) = (4x(x^2-1)^2(x^2+2))/(x^2+1)^2#
To find the second derivative , one method is to multiply out the numerator of the first derivative and then use the quotient rule:
#f'(x) = (4x(x^4-2x^2+1)(x^2+2))/(x^2+1)^2#
#f'(x) = ((4x^5-8x^3+4x)(x^2+2))/(x^2+1)^2#
#f'(x) = (4x^7 + (8x^5) - 8x^5 -16x^3+4x^3+8x)/(x^2+1)^2#
#f'(x) = (4x^7-12x^3+8x)/(x^2+1)^2#
Let #u =4x^7-12x^3+8x = 4x(x^6-3x^2+2)#
#u' = 28x^6-36x^2+8 = 4(7x^6-9x^2+2)#
Let #v = (x^2+1)^2, v' = 2(x^2+1)(2x) = 4x(x^2+1)#
#f''(x) = #
#((x^2+1)^2*4(7x^6-9x^2+2) - 4x(x^6-3x^2+2)4x(x^2+1))/ (x^2+1)^4#
Factor:#f''(x) =#
# (4(x^2+1)[(x^2+1)(7x^6-9x^2+2)-4x^2(x^6-3x^2+2)])/(x^2+1)^4#
Cancel one if the #(x^2+1)# from both numerator & denominator:
#f''(x) = (4[(x^2+1)(7x^6-9x^2+2)-4x^2(x^6-3x^2+2)])/(x^2+1)^3#
Simplify the numerator through distribution:
#f''(x) = #
#(4[7x^8-9x^4+2x^2+7x^6-9x^2+2-4x^8+12x^4-8x^2])/(x^2+1)^3#
Simplify:
#f''(x) = (4(3x^8+7x^6+3x^4-15x^2+2))/(x^2+1)^3#
