Second differential of function

Question

I have the following function: $y = u(2+v)$, also I know $, ^2, , ^2$. I need to find the second differential $d^2y$. I understand, that this not similar to finding second derivative, but I do not know the direction in which to move. I used definition $df(x) = f'(x)dx$.

Answer 1

It depends on what book you are using. I calculate second differentials in a way that is algebraically manipulable. Using that approach, you can find the first differential as follows:

$$y = u(2 + v)\\ y = 2u + uv \\ d(y) = d(2u + uv) \\ d(y) = d(2u) + d(uv) \\ dy = 2du + u\,dv + v\,du \\ $$

The second differential is just $d(d(y))$ or $d(dy)$ or $d^2y$ (all the same thing). So we just apply the differential again:

$$ dy = 2du + u\,dv + v\,du \\ d(dy) = d(2du + u\,dv + v\,du) \\ d(dy) = d(2du) + d(u\,dv) + d(v\,du) \\ d^2y = 2d^2u + u\,d^2v + dv\,du + v\,d^2u + dv\,du \\ d^2y = 2d^2u + u\,d^2v + 2\,dv\,du + v\,d^2u $$

Answer 2

Differentiating with respect to an arbitrary variable $t$ and dropping off $dt, d^2t$ as in your notation retaining differentials only, we can adopt an abbreviated notation

$$ u'= du, u''=d^2u,\; v'=dv,v''=d^2v$$ $$ y= 2u+uv,\; y'=2u'+u'v+uv'\;$$ differentiate once again, (reminds binomial power expansion) $$y''= 2u''+\; u''v+2 u'v'+uv''. $$