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Is this the right way of solving $frac{d}{dx}(sin(x)cdot x^2)$

Mathematics Asked on November 1, 2021

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 $dx^2$.

Figure - 1

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.
$$therefore~ d(sin(x)cdot x^2) = sin(x)cdot dx^2+x^2cdot d(sin(x))+dx^2cdot d(sin(x))$$.
Now, $dx^2 = 2xcdot dx$, $d(sin(x)) = cos(x)cdot dx$.
$$therefore~ dfrac{d(sin(x)cdot x^2)}{dx} = dfrac{sin(x)cdot dx^2+x^2cdot d(sin(x))+dx^2cdot d(sin(x))}{dx}$$
$$ = dfrac{sin(x)cdot 2xcdot dx + x^2cdotcos(x)cdot dx + 2xcdot dxcdot cos(x)cdot dx}{dx}$$
$$ = sin(x)cdot 2x + x^2cdot cos(x) + 2xcdot cos(x)cdot dx$$
Now, as $dx rightarrow 0$, $dfrac{d(sin(x)cdot x^2)}{dx} rightarrow sin(x)cdot 2x + x^2cdotcos(x)$ 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 $dfrac{d(sin(x)cdot x^2)}{dx} = sin(x)cdot 2x + x^2cdotcos(x)$ 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!

One Answer

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.

Answered by user809793 on November 1, 2021

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