## anonymous 5 years ago 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

1. anonymous

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

2. anonymous

$f(x)=g(e ^{2x})$ $df/dx=d(g(e ^{2x})/dx$ 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

3. anonymous

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

4. anonymous

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

5. anonymous

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.