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

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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

OCW Scholar - Multivariable Calculus
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so I suppose you are trying to find f''(x)
\[f(x)=g(e ^{2x})\] \[df/dx=d(g(e ^{2x})/dx\] 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
\[f''(x)=d ^{2}f/dx ^{2}=d ^{2}g/du ^{2} \times (du/dx)^{2}+dg/du \times d ^{2}u/dx ^{2}\]

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

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