Answer
$$f''(x)=6xg'(x^2)+4x^3g''(x^2)$$
Work Step by Step
$$f(x)=xg(x^2)$$
1) First derivative of $f(x)$ $$f'(x)=x'g(x^2)+x[g(x^2)]'$$ $$f'(x)=g(x^2)+x[g(x^2)]'$$
- Considering $[g(x^2)]'$ $$[g(x^2)]'=\frac{dg(x^2)}{dx}$$
Apply the Chain Rule: $$[g(x^2)]'=\frac{dg(x^2)}{d(x^2)}\frac{d(x^2)}{dx}$$ $$[g(x^2)]'=g'(x^2)\times2x$$ $$[g(x^2)]'=2xg'(x^2)$$
Therefore, $$f'(x)=g(x^2)+2x^2g'(x^2)$$
2) Second derivative of $f(x)$ $$f''(x)=[g(x^2)+2x^2g'(x^2)]'$$ $$f''(x)=[g(x^2)]'+2[(x^2)'g'(x^2)+x^2[g'(x^2)]']$$
- Considering $[g'(x^2)]'$ $$[g'(x^2)]'=\frac{dg'(x^2)}{dx}=\frac{dg'(x^2)}{d(x^2)}\frac{d(x^2)}{dx}$$ (Chain Rule) $$[g'(x^2)]'=2xg''(x^2)$$
- Also, from part 1), we prove that $[g(x^2)]'=2xg'(x^2)$
- Also, $(x^2)'=2x$
Therefore, $$f''(x)=2xg'(x^2)+2[2xg'(x^2)+2x^3g''(x^2)]$$ $$f''(x)=2xg'(x^2)+4xg'(x^2)+4x^3g''(x^2)$$ $$f''(x)=6xg'(x^2)+4x^3g''(x^2)$$
