Derivatives and their proofs

Question

I have seen in many videos and in some books the following proof: \begin{align*} y &= x^2 \\ \implies y + dy &= (x+dx)^2\\ \implies y+dy &= x^2 + (dx)^2 +2x\,dx\\ \end{align*} Substituting $y = x^2$, \begin{align*} x^2 + dy &= x^2 + (dx)^2+2x\,dx \\ \implies dy &= (dx)^2 +2x\,dx\\ \end{align*} Neglecting $(dx)^2$ as it is very small, $$\frac{dy}{dx}=2x$$

I don't understand how we can neglect the $(dx)^2$ even if it is very small, that seems like an approximation and not an exact answer. Also, can we even write $dx$ separately like this, isn't $\frac{dy}{dx}$ just notation.

Answer 1

• For the first part of your question-

Try imagining $f(x)dx$ as a rectangle with with length $f(x)$ and breadth $dx$. $dx$ being very small when you integrate you add the very 'insignificant' amount of areas together to get the total area of a graph (for example - like how your book has very large thickness compared to the paper). So when you write $(dx)^2$, here both the length and the breadth become insignificant and hence can be ignored. For a more visual argument look at the picture attached.

a visual argument In the image the green area is, as in your example, $x^2$ and the area of the the entire square is $(x+dx)^2$ notice how the area of the red square i.e. $(dx)^2$ will become insignificant as $dx$ tends to $0$.

• For your second question-

$dx$ & $dy$ are often used as a ratio as they are dimentionally correct and do not give any contradiction (not that i know of). If you want to be more rigorous about it you can use $Δx$ and let the limit tend to zero.

Mathematically $\frac{d()}{dx}$ is a symbol in itself, separating the numerator and denominator will be like separating the two lines in the symbol of addition. Same goes for integration : $∫()dx$ is a symbol in itself , you can not separate the $dx$, but will be allowed in 'non-rigorous' courses as again it gives no contradiction and you will end up getting the correct answer.

Answer 2

This is what I would consider a non-rigorous way of introducing the derivative. The complete, rigorous approach requires a formal definition of limits.

Without going too deep on the topic, I would offer the following idea (which is based on the formal definition but still presented casually):

• If we have something that we can approximate, and if we have a way of making the error of that approximation as small as we like by measuring things on an increasingly small scale, then by taking that approximation error down to zero we get the actual value of the thing we're measuring.

So in other words, we have this expression $\frac{\delta y}{\delta x} = 2x + \delta x$ which is true for any value of $\delta x$ except possibly zero. We are saying that $\frac{dy}{dx} \approx \frac{\delta y}{\delta x}$, meaning that the derivative is approximately equal to this other expression, and when $\delta x$ is small we expect the error on the approximation to also be small. So if we make $\delta x$ arbitrarily small, we can get $\frac{\delta y}{\delta x}$ arbitrarily close to being equal to $2x$, and so we define the limit, as $\delta x$ approaches zero, to be exactly this value.

Note that here I made a distinction between the "approximation" using $\delta x$ and $\delta y$, and the "limit" using $dx$ and $dy$. Not distinguishing between the two is one of the aspects that makes the approach you've shown less rigorous, although there are systems where that kind of manipulation is perfectly fine. It's also important that the "error" term in the approximation has to shrink as $\delta x$ shrinks - terms like $\delta x, (\delta x)^2, (\delta x)^{100}$ all vanish when $\delta x$ vanishes, but $\frac{1}{\delta x}$ gets arbitrarily large as $\delta x$ gets arbitrarily small, so if after cancelling everything out you have a term like that then you definitely can't expect the limit to work properly.