First we need to use the product rule, as follows:
Differentiating Once:
# f(x)=xg(x^2) #
# :. f'(x)=(x)(d/dxg(x^2)) + (d/dxx)(g(x^2))#
# :. f'(x)=x(d/dxg(x^2)) + (1)(g(x^2))#
# :. f'(x)=x(d/dxg(x^2)) + g(x^2)#
Then By the chain rule, we have:
# d/dxg(x^2) = g'(x^2)(d/dxx^2) #
# d/dxg(x^2) = g'(x^2)(2x) #
# :. d/dxg(x^2) = 2xg'(x^2) # .... [1]
Substituting [1] gives us:
# :. f'(x)=x(2xg'(x^2)) + g(x^2)#
# :. f'(x)=2x^2g'(x^2) + g(x^2)#
Differentiating a second time, (again using the product rule):
# f''(x) = (2x^2)(d/dxg'(x^2)) + (d/dx2x^2)(g'(x^2)) + (d/dxg(x^2))#
Again, Substituting [1] gives us:
# f''(x) = (2x^2)(d/dxg'(x^2)) + (4x)(g'(x^2)) + (2xg'(x^2))#
# f''(x) = 2x^2(d/dxg'(x^2)) + 6xg'(x^2)#
And, again by the chain rule, we have:
# d/dxg'(x^2) = (g''(x^2))(d/dxx^2) #
# :. d/dxg'(x^2) = (g''(x^2))(2x) #
# :. d/dxg'(x^2) = 2xg''(x^2) # .... [2]
Substituting [2] gives us:
# f''(x) = 2x^2(2xg''(x^2)) + 6xg'(x^2)#
# f''(x) = 4x^3g''(x^2) + 6xg'(x^2)#
