0
$\begingroup$

If $x=e^t$, can someone give me proof that $$ \frac{d^2}{dx^2}=\frac{1}{e^{2t}}\left(\frac{d^2}{dt^2}−\frac{d}{dt}\right). $$ Thank you

$\endgroup$

closed as off-topic by caverac, RRL, Saad, José Carlos Santos, mrtaurho Dec 27 '18 at 12:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – caverac, RRL, Saad, José Carlos Santos, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

$$ \frac{d}{dx} = \frac{d}{dt} \times \frac{dt}{dx} = \frac{d/dt}{dx/dt} = e^{-t} \frac{d}{dt}. $$

Note that $$ \frac{d^2}{dx^2} = \frac{d}{dx} \left[ \frac{d}{dx} \right] $$ and you can now use the previous result.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.