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I was trying to prove

$$d(xy) = x(dy) + y(dx)$$

earlier this morning and I used this post to help me understand the task.

I understood the entirety of the post for my calculus class, apart from one step.

Considering an area of a rectangle with dimensions $x$ and $y$ $xy$ makes sense.

Likewise, the area of a rectangle with dimensions $(x+\Delta x)(y+\Delta y)$ giving

$$A_1=(x+\Delta x)(y+\Delta y)=xy+x \Delta y+y\Delta x+\Delta x \Delta y$$

made sense too.

I was confused however about the subtraction of the two areas that gives

$$x \Delta y+y\Delta x+\Delta x \Delta y$$

but the small approximation of $\Delta x $ and $\Delta y$ very small, ensuring that $\Delta x \Delta y$ is negligible afterwards made sense.

Why is there the need to subtract the areas as part of the proof for this differentiation property?

vik1245
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    You are concerned with the change in area of the rectangle, i.e., the new area minus the old area. Hence you subtract $xy$. – kccu Mar 31 '19 at 17:34
  • By definition, $\Delta(xy)=(x+\Delta x)(y+\Delta y)-xy$, and $d(xy)$ is the linear approximation of $\Delta(xy)$. – Eclipse Sun Mar 31 '19 at 17:39
  • $xy$ is the current area. If you are concerned with the change, you look only at the terms with $\Delta$. Once we talk about infinitesimal changes, we get differentials. – Andrew Li Mar 31 '19 at 18:14
  • $dx$ is not a rigorous quantity – Vivaan Daga Jun 26 '21 at 09:03

2 Answers2

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By $d(xy)$ we mean change in xy, Remember when a differential was defined, it was "Infinitesimal Change".

  • Wikipedia says:

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What exactly is change? Roughly speaking Final - Initial.

The initial here is $xy$.

We wish to express final in terms of change in the individual variables $x$ and $y$.

Hence we do $$d(xy) = (x+dx)(y+dy) - xy \ \ \ [Final \ - \ Initial]$$

RishiNandha Vanchi
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Whenever dealing with derivatives change is an inevitable part. Now , what the person was trying to explain is let $A=x.y$ and now he was trying to find the CHANGE In Area $A$ for some infinitesimal small change in an arbitrary variable z. Mathematically that means $\frac{dA}{dz}$ which is similar to $\frac{\Delta A}{\Delta z}$ when the change in $\Delta z$ is not small and as we know $\Delta A $ means change in area of triangle i.e $A_0 -A_1$ which is what is done in the post .

Now , $\frac{dA}{dz}=\frac{d(x.y)}{dz}= x.\frac{dy}{dz}+y.\frac{d x}{dz}$(product rule) Taking ,$dz$ common from both sides and eliminating gives $d(x.y)=x.dy+y.dx$

Soumya Patel
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