7
$\begingroup$

I've seen $d(xy) = y(dx) + x(dy)$, but I don't understand the principle behind it and memorizing it is lame. Can anyone explain what is going on here?

For example from physics, $$F = {{dP} \over {dt}}$$ $$F = {{d(mv)} \over {dt}}$$ $$F = {{v(dm) + m(dv)} \over {dt}}$$ Since $$dm = 0$$

$$F = {{m(dv)} \over {dt}}$$ Since $$a = {{dv} \over {dt}}$$ $$F = ma$$

$\endgroup$
8
  • $\begingroup$ Are you looking for intuition, or proofs? (Ideally you'd want a highly intuitive proof, but that's not always available.) $\endgroup$ – user2357112 supports Monica Jun 10 '14 at 8:55
  • $\begingroup$ What do you mean by "principle"?? $\endgroup$ – tattwamasi amrutam Jun 10 '14 at 8:59
  • $\begingroup$ it is the product rule found by Leibniz. Look at the principle here: en.wikipedia.org/wiki/Product_rule $\endgroup$ – Martial Jun 10 '14 at 9:05
  • 8
    $\begingroup$ $$\Delta(xy)=(x+\Delta x)(y+\Delta y)-(xy)\approx x\ \Delta y+y\ \Delta x.$$ $\endgroup$ – Yves Daoust Jun 10 '14 at 9:38
  • $\begingroup$ You're right, I better just memorize it as Leibniz is, for the moment, beyond my mind's intellectual capacity. He was a very good mathematician I understand. Whether Newton or Leibniz developed the calculus first is an irrelevant question in my opinion. $\endgroup$ – Michael Lee Jun 10 '14 at 18:54
13
$\begingroup$

Hint

Suppose a rectangle of dimensions $x$ and $y$; its area is $A_0=xy$. Now change $x$ to $x+\Delta x$ and $y$ to $y+\Delta y$. The area of the new rectangle is given by $$ A_1=(x+\Delta x)(y+\Delta y)=xy+x \Delta y+y\Delta x+\Delta x \Delta y$$ So, the change of the area is $$\Delta A=A_1-A_0=x \Delta y+y\Delta x+\Delta x \Delta y$$ Now, let us make $\Delta x $ and $\Delta y$ very small; then $\Delta x \Delta y$ is negligible.

I am sure that you can take from here.

In order to illustrate, let us consider $x=10$ meters, $y=5$ meters and $\Delta x=\Delta y=1$ centimeter that is to say $0.01$ meters. So, using the last formula, $$\Delta A=10 \times 0.01+5 \times 0.01+0.01\times 0.01=0.1501$$ You see how small is the last term compared to the previous ones.

$\endgroup$
2
  • 1
    $\begingroup$ Now I see Claude. $dxdy$ is getting much smaller per decrement of $dx$ and $dy$. In other words, $ydx + xdy + dxdy \approx ydx + xdy$ $\endgroup$ – Michael Lee Jun 13 '14 at 12:40
  • $\begingroup$ Bingo ! You made it ! Now, you must go from $\Delta$'s to $\delta$'s and when these ones become infinitesimal, then you get what you wrote. I am glad to have been able to help you. Cheers :) $\endgroup$ – Claude Leibovici Jun 13 '14 at 16:01
1
$\begingroup$

Here you go: https://math.stackexchange.com/a/3393408/633100

I'll actually Paste the same answer here (I tried sharing link alone but apparently answers have to be longer than 30 characters)

$d(xy)=(x+dx)(y+dy)-xy \ $ [from Definition of a "differential"]

$\ \ \ \ \ \ \ \ \ =ydx + xdy + dxdy $

Here what people normally do is, "$dxdy$ is too small and negligible". But that is SO BAD.

Instead, go back to definition of the derivative.

$\lim_{ h -> 0 }\frac{f(x+h)+f(x)}{h}$

Divide the first equation we got by $dx$

$(xy)' = y + x(y') + dy $

$dy -> 0$ by definition of a derivative/a differential.

Another way of looking at this would be, Integrate both sides and all the terms will become bigger values except this $dxdy$ will alone have to be integrated twice to give it a good value.

for any quantity z, dz is defined as the infinitesimally small change in that quantity. ie approaches 0. These are generally used in combination with one or more 'differentials' and so we normally don't cancel of these to equal 0. for instance, $\frac{dv}{dt}$ is not 0/0. But with $dxdy$ if you integrate the whole equation, all the terms of the equation move from being differentials to regular terms but except this last one. There are no other differentials in the equation which makes it legal to cancel the last one to 0.

At least that's how I see it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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