What is meant by $d(xy)$?

Question

$dy/dx$ means derivative of $y$ with respect to $x$ But what is meant by $d(xy)$? Where is the "with respect to term" here ?

Answer 1

$dx$ can be thought of as an infinitely small change in $x$, just as $\Delta x$ means a change in $x$ that is not infinitely small, and $dy$ would be the resulting infinitely small change in $y$. Recall that $$ \frac{dy}{dx} = \lim_{\Delta x\to0} \frac{\Delta y}{\Delta x}. $$

Likewise $d(xy)$ would be an infinitely small change in the product $xy$. If $x$ and $y$ are both functions of $t$, then the product rule can be stated as $$ \frac{d(xy)}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}, $$ and sometimes it is actually written as $$ d(xy) = x\,dy + y\,dx. $$ It can be rearranged into this: $$ y\,dx = d(xy) - x\,dy $$ and then both sides can be integrated: $$ \int y\,dx = xy - \int x\,dy. $$ In that form, it is called integration by parts.

Answer 2

It is the differential of $xy$ which is $$ d(xy)=x\,dy+y\,dx $$ As for a somewhat intuitive interpretation, noticing that $$ \Delta(xy)=(x+\Delta x)(y+\Delta y)-(xy)=x\Delta y+y\Delta x+\Delta x\Delta y\approx x\ \Delta y+y\ \Delta x. $$ you can think of it as giving the increment in $xy$ in terms of the increments of $x$ and $y$, when those increments become infinitesimal (so that the second order terms become negligible).

Answer 3

It is called the differential of $xy$, and denotes the linear approximation of the increment of $xy$ when $x$ and $y$ have small increments $h$ and $k$: $$\Delta(xy)=(x+h)(y+k)-xy=\underbrace{xh+yk}_{\text{linear part in $h,k$}}+hk=\mathrm d\mkern1mu(xy)(h,k)+hk.$$