The first derivative of a product of two polynomials Consider the differentiation of functions which are a product of two polynomials, such as $y = (x^2-1)(x+1)$
Let u = $x^2-1$, and v = $x+1$
Then, $y=uv$ ........................(1)
where u and v are both functions of x
From the first principles of differentiation:
$y+{\partial y} = (u + {\partial u})(v +{\partial v})$
= $uv + u{\partial v} + v{\partial u} + {\partial u}{\partial v}$ ........................(2)
(2)-(1):
${\partial y} = u{\partial v} + v{\partial u} + {\partial u}{\partial v}$ ........................(3)
Divide (3) by ${\partial x}$:
$\frac{\partial y}{\partial x}$ = $u \frac{\partial v}{\partial x} + v \frac{\partial u}{\partial x} + \frac {\partial u \partial v}{\partial x}$
Take the limits on both sides:
$\lim_{\partial x\to 0} \frac {\partial y}{\partial x} = u \lim_{\partial x\to 0} \frac {\partial v}{\partial x} + v \lim_{\partial x\to 0} \frac {\partial u}{\partial x} + \lim_{\partial x\to 0} \frac {\partial u}{\partial x} {\partial v} $
As ${\partial x\to 0}$, ${\partial v}$ in general also tend to zero.
$\lim_{\partial x\to 0} \frac {\partial y}{\partial x} = u \lim_{\partial x\to 0} \frac {\partial v}{\partial x} + v \lim_{\partial x\to 0} \frac {\partial u}{\partial x} + \lim_{\partial x\to 0} \frac {\partial u}{\partial x} (0)$
$\lim_{\partial x\to 0} \frac {\partial y}{\partial x} = u \lim_{\partial x\to 0} \frac {\partial v}{\partial x} + v \lim_{\partial x\to 0} \frac {\partial u}{\partial x}$
Since $\lim_{\partial x\to 0} \frac {\partial y}{\partial x} = \frac {dy}{dx}$, hence, $\frac {\partial v}{\partial x}\to \frac {dv}{dx} and \frac {\partial u}{\partial x}\to \frac {du}{dx}$
Generally,
$$ \frac {d}{dx} (uv) = u \frac {dv}{dx} + v \frac {du}{dx}$$
This is known as the product rule for differentiation.
Everything that I wrote up there is the exact copy of what is written on my school textbook.
My question here is, when it said
"As ${\partial x\to 0}$, ${\partial v}$ in general also tend to zero",
how come only the $\lim_{\partial x\to 0} \frac {\partial u}{\partial x} {\partial v} $ is affected, making it $\lim_{\partial x\to 0} \frac {\partial u}{\partial x} (0)$? How come the $u \lim_{\partial x\to 0} \frac {\partial v}{\partial x}$ is not affected too, since it also has the value $\partial v$
Although I know how to use the equation in questions, I'm just curious on how they derive the formula. I am aware that the other way of solving this question would be by expanding the function and finding its $ \frac {dy}{dx}$ from there.
I've also tried to look for other sources but non of them seem to really answer my question. My older brother told me it's unnecessary that I understand the whole process behind it since I'm still in high school and that I only need to know how to correctly apply the formula when solving questions. He did say that if I do insist on understanding it in depth then that's for when I do a Master's degree? I hope by posting this question I don't confuse myself even more.
Thanks!
Edit: Sorry, here's a picture of the whole thing instead:  http://imgur.com/a/i628E
 A: The crucial observation here is that $\lim_{\delta x\rightarrow 0}\frac{\delta v}{\delta x}=\frac{0}{0}$, an indeterminate form. Note that continuous functions will always (by definition) have $\delta v\rightarrow 0$ as $\delta x\rightarrow 0$. 
Without more information about the function itself, it is impossible to know what happens in this case, though if the function is differentiable (which is usually the case in calculus and precalc courses), by definition this indeterminate form has a finite value.
In the case of the $\lim_{\delta x\rightarrow 0}\frac{\delta u}{\delta x}\delta v$, let $\lim_{\delta x\rightarrow 0}\frac{\delta u}{\delta x}=\alpha$. Then $$\lim_{\delta x\rightarrow 0}\frac{\delta u}{\delta x}\delta v = \lim_{\delta x\rightarrow 0}\frac{\delta u}{\delta x}\lim_{\delta x\rightarrow 0}\delta v = \alpha \lim_{\delta x\rightarrow 0} \delta v = \alpha \cdot 0=0$$
We can break up the limit of the product to the product of the limits in the first step due to properties of limits.
To contrast with the other case, if we let $\lim_{\delta x\rightarrow 0}\frac{\delta v}{\delta x}=\beta$ we get
$$u\lim_{\delta x\rightarrow 0}\frac{\delta v}{\delta x}=u\cdot \beta$$ which could be a nonzero quantity.
A: Your older brother is probably right that if you just want to pass your class, you don't need to understand everything perfectly. But it sounds like you want to, and there's absolutely no reason you can't attain a really good grasp of Calculus even in high school, if you're willing to put in the effort. 
Let me address the second part of your question first: Why doesn't $u\lim_{\delta x\to0}\frac{\delta v}{\delta x}$ go to $0$? 
As $\delta x$ goes to $0$, it's certainly true that $\delta v$ goes to $0$, so at first glance, we might want to say that the whole thing goes to $0$. But since the denominator goes to $0$ as well, that is not necessarily true. We have a very small quantity divided by another very small quantity, and as they both approach $0$, the ratio can approach something nonzero. In fact, the $\frac{\delta v}{\delta x}$ approaches $\frac{dv}{dx}$. Since this is multiplied by $u$, that whole term approaches $u\frac{dv}{dx}$. Similarly with the second term.
Now what is the difference in the third term? Recall the product rule for limits: the limit of a product is the product of the limits (contrast this with the derivative of a product, which is NOT the product of the derivatives, as this very discussion shows). So then: $\lim_{\delta x\to0}\frac{\delta u}{\delta x}\delta v=\lim_{\delta x\to0}\frac{\delta u}{\delta x}\cdot\lim_{\delta x\to0}\delta v=\frac{du}{dx}\cdot0$.
What happened here? The $\delta v$ was not divided by another small quantity, since the $\delta x$ was busy keeping the $\delta u$ in check. So the $\delta v$ just goes to $0$, bringing the whole term to $0$. Note that we could have also written this term as $\delta u\cdot\frac{\delta v}{\delta x}$. In that case, the $\frac{\delta v}{\delta x}$ would have gone to $\frac{dv}{dx}$, but the $\delta u$ would have gone to $0$. The point is that the $\delta x$ can only take care of one thing at a time.
Another way to look at it is this: Suppose $\frac{du}{dx}=5$. (or any other number). Eventually, with $\delta x$ small enough, we will have $\frac{\delta u}{\delta x}$ really close to $5$. Say we make $\delta x$ small enough that $\frac{\delta u}{\delta x}$ is between $4.9$ and $5.1$. Then we can also make $\delta x$ small enough that $\delta v$ is close to $0$. Say we make $\delta x$ small enough that $\delta v$ is between $-.00001$ and $.00001$. Then $\frac{\delta u}{\delta x}\delta v$ is at most $5.1$ times $.00001$, which is really close to $0$. You can probably see how making $\delta x$ even smaller makes this quantity even closer to $0$. In fact, you can make the quantity as close to $0$ as you want, just by making $\delta x$ small. And that's exactly what it means for the limit of the whole term to be $0$.
