Is $\frac{\Delta y}{\Delta x}$ equal to $f'(x) + \varepsilon$? 
I was trying to understand the relationship between differential and derivative, but I couldn't understand where the first equation comes from.
When,
$$\Delta y = f(x + \Delta x) - f(x),\ \ \ \ \lim_{\Delta x \to 0} \varepsilon = 0$$
Is this equation true?
$$
\frac{\Delta y}{\Delta x} = f'(x) + \varepsilon
$$
 A: What you are essentially asking is whether or not it is true that, if $f$ is differentiable at $x$ then the difference
$$\frac{f(x+\Delta x)-f(x)}{\Delta x}-f'(x)$$
tends to zero as $\Delta x$ tends to zero,  and the answer is, yes, it does, because that is precisely the definition of the derivative.
However, it is not true that the difference above can be guaranteed to be equal to an arbitrarily pre-assigned function $\varepsilon$ that satisfies the condition $\lim_{\Delta x\to0}\varepsilon=0$. For example, the function $\varepsilon(x)=x^2$ certainly tends to zero when $x\to 0$, but with $f(x)=x^2$, we have
$$\frac{\Delta y}{\Delta x}=\frac{(x+\Delta x)^2-x^2}{\Delta x}=2x+\Delta x$$
and we know that $f'(x)=2x$, so 
$$\frac{\Delta y}{\Delta x}-f'(x)=\Delta x\neq\varepsilon(x)$$
A: Firstly, an equation cannot "exist", so your terminology that the "equation exists" is rather weird, to say the least. It's important here to understand really what you're asking, and to do tbat you must first understand what the equation means. Intuitively, the statement $$\frac{\Delta y}{\Delta x}=f'(x)+\epsilon$$ says that if $\Delta x$ is small, then $\Delta y/\Delta x$ (this is the gradient of the straight secant line approximation to the tangent line at a point $x$) is approximately the same as $f'(x)$, in the sense that it is at most $\epsilon$ more than $f'(x)$, where $\epsilon$ here is a small real number. If $f$ is differentiable, intuitively you can see this by taking a point $x$ and drawing tangent and secant lines through that point. What the equation really says is that $f'(x)$ and $\Delta y/\Delta x$ are "close".
Rigorously speaking, what this really says is that for every $\epsilon>0$, there exists a $\Delta x>0$ so that $|\Delta y/\Delta x-f'(x)|<\epsilon$ for some $\Delta x$ which we can choose; or equivalently, that $$\lim_{\Delta x\to0}\left(\frac{\Delta y}{\Delta x}-f'(x)\right)=0.$$
Again, you should understand this statement as the claim that $\Delta y/\Delta x$ gets arbitrarily clsoe to $f'(x)$ as $\Delta x\to 0$. Of course, this only holds if $f$ is differentiable.
