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I don't know much about theorems related to limits and I'm currently learning calculus from 3Blue1Brown's online series : Essence of Calculus.

This example's generalization is what I'll use to derive the product rule.

Now, if we have a rectangle with length and breadth equal to $\sin(x)$ and $x^2$ for some values of $x$, then it's area will be : $\sin(x)\cdot x^2$. Now, if we "nudge" the value of $x$ by some little amount, say, $dx$, then there will be corresponding changes in the values of $\sin(x)$ and $x^2$. Let the little change in $\sin(x)$ be $d(\sin(x))$ and the little change in $x^2$ be $dx^2$.

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Now, the change in the value of $(\sin(x)\cdot x^2)$ i.e. $d(\sin(x)\cdot x^2)$ will be the sum of the three new strips of area. $$\therefore~ d(\sin(x)\cdot x^2) = \sin(x)\cdot dx^2+x^2\cdot d(\sin(x))+dx^2\cdot d(\sin(x))$$. Now, $dx^2 = 2x\cdot dx$, $d(\sin(x)) = \cos(x)\cdot dx$. $$\therefore~ \dfrac{d(\sin(x)\cdot x^2)}{dx} = \dfrac{\sin(x)\cdot dx^2+x^2\cdot d(\sin(x))+dx^2\cdot d(\sin(x))}{dx}$$ $$ = \dfrac{\sin(x)\cdot 2x\cdot dx + x^2\cdot\cos(x)\cdot dx + 2x\cdot dx\cdot \cos(x)\cdot dx}{dx}$$ $$ = \sin(x)\cdot 2x + x^2\cdot \cos(x) + 2x\cdot \cos(x)\cdot dx$$ Now, as $dx \rightarrow 0$, $\dfrac{d(\sin(x)\cdot x^2)}{dx} \rightarrow \sin(x)\cdot 2x + x^2\cdot\cos(x)$ because anything in the form of $p(dx)^n$, where $p \in \Bbb R$ and $n \in \Bbb Z^+$ will approach $0$ as well.

So, we can say that $\dfrac{d(\sin(x)\cdot x^2)}{dx} = \sin(x)\cdot 2x + x^2\cdot\cos(x)$ as $$\dfrac{d}{dx} f(x) = \lim_{\Delta x \rightarrow 0}\dfrac{f(x+\Delta x) - f(x)}{\Delta x}$$

I want to know if I've done all of this correctly, without any conceptual mistakes.

Thanks!

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    $\begingroup$ As a heuristic this is alright. If you want to make these kind of arguments rigourous, you might learn synthetic calculus, as it is done in John L. Bell's A primer of infinitesimal analysis. $\endgroup$ – Jackozee Hakkiuz Jul 19 '20 at 0:26
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    $\begingroup$ To follow that book you'd need to feel comfortable with reading and writing proofs and have a certain familiarity with abstract algebra, so it may not be appropriate just yet. If you go and take a rigourous calculus course in school it will surely be based on limit techniques. That approach uses a different kind of reasoning, but it works fine and you get the same results. $\endgroup$ – Jackozee Hakkiuz Jul 19 '20 at 0:51
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    $\begingroup$ If your question is whether your derivation is correct in the usual mathematical framework, then the answer is no, because in that context $\frac{df}{dx}$ is not a fraction but a formal symbol: in particular, you cannot "divide by $dx$". However, in the context of 3B1B videos, it's all good. $\endgroup$ – Jackozee Hakkiuz Jul 19 '20 at 0:52
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    $\begingroup$ You're welcome. Keep following your interests. If you want to get into calculus formally, an excellent book is "Calculus" by Michael Spivak. All the intuition you already have thanks to 3B1B will surely be useful. $\endgroup$ – Jackozee Hakkiuz Jul 19 '20 at 0:56
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    $\begingroup$ @Pleasecorrectgrammarmistakes If you want the link to his series on calculus, here it is : youtube.com/… $\endgroup$ – Rajdeep Sindhu Jul 20 '20 at 2:30
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You forgot, not just one time, that $\dfrac{dx(\sin(x).x^2)}{dx}\to \sin(x).2x + x^2.\cos(x) $ and not $dx(\sin(x).x^2) \to \sin(x).2x + x^2.\cos(x)$.

You can't define $n > 0$, because if $n = 0$ so anything in the form $p(dx)^0$ will approach p.

If you wanna study calculus, I recommend using a Calculus book or videos on internet, like MIT Single Variable Calculus. 3Blue1Brown makes incredible videos to make things look more interesting, but you can't use his definitions to solve something like $f(x) = \frac{|x|}{x}$ and $g(x) = |x|+ x$, without understanding $\frac{d(|x|)}{dx}$. Because, what means that something is not defined when approaches to 0?

I use Calculus by Michael Spivak on my course, it's pretty good. But for a first contact with Calculus, any book will be great, you could understand more things studying Real Analysis later, for example.

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    $\begingroup$ I have fixed the error that you pointed out at the start of your post, it was a typo. I don't quite get what you mean when you say that You can't define n>0, because if n = 0, so anything in the form p(dx)^0 will approach p. I stated that $n \in \Bbb N$, which means that $n > 0$. Did I misunderstand something that you stated? Finally, thank you for your recommendations. $\endgroup$ – Rajdeep Sindhu Jul 19 '20 at 1:28
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    $\begingroup$ For me and much other people, 0 is natural. But some mathematics don't define that 0 is natural. So for better understanding define that n is a positive integer (n > 0) $\endgroup$ – user809793 Jul 19 '20 at 1:41

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