4 ms·
>For functions of a single parameter, /partial operator/ is equivalent to /full derivative operator/ (for sufficiently smooth functions). Are you pulling my le
by letlambda 9y ago
>For functions of a single parameter, /partial operator/ is equivalent to /full derivative operator/ (for sufficiently smooth functions).
Are you pulling my leg here, or do I need to scrap my understanding of calculus?
If they exist, how could they not be equivalent?