how would I calculate the double prime of F for f(x) = g(e^2x), where g is a function defined for all real numbers and g admits second order derivative

Question

anonymous · Answer

so I suppose you are trying to find f''(x)

anonymous · Answer

$$f(x)=g(e ^{2x})$$ $$df/dx=d(g(e ^{2x})/dx$$ let $$e ^{2x}=u$$ then by chain rule $$df/dx=dg/du \times du/dx$$ and $$d ^{2}f/dx ^{2}=d(dg/du \times du/dx)/dx$$ $$=d(dg/du)/dx \times du/dx+dg/du \times d ^{2}u/dx ^{2}$$ but $$d(dg/du)/dx=d(dg/du)/du \times du/dx=d ^{2}g/du ^{2} \times du/dx$$ then

anonymous · Answer

$$f''(x)=d ^{2}f/dx ^{2}=d ^{2}g/du ^{2} \times (du/dx)^{2}+dg/du \times d ^{2}u/dx ^{2}$$

anonymous · Answer

I hope this is correct, any suggestions are more than welcome.

anonymous · Answer

chain rule
$$f \prime(x)=g \prime (e ^{2x})2e^{2x}$$
product rule
$$f \prime\prime (x)=(g \prime (e ^{2x})2e^{2x})\prime=g \prime\prime  (e ^{2x})4e^{4x}+g \prime  (e ^{2x})4e^{2x}$$
which is the same thing and i also hope is correct.