Confusing differential concept for $f(x)$ and $f(x,y)$.

Question

Okay, so I was reading the concept of "Increment and Differential of a Function" from the book "Mathematics: Its Contents, Methods and Meaning".
It says: Let us consider a function $y=f(x)$ which has a derivative. The increment of this function $$\Delta y=f(x+\Delta x)-f(x)$$ corresponding to the increment $\Delta x$, has the property that the ratio $\frac{\Delta y}{\Delta x}$ approaches a finite limit as $\Delta x\to 0$ equal to the derivative. The fact may be written as an equality $\frac{\Delta y}{\Delta x}=f'(x)+\alpha$, where the value of $\alpha$ depends on $\Delta x$ in such a way that as $\Delta x\to0$, it also approaches $0$. Thus increment of a function may also be represented as $$\Delta y=f'(x)\Delta x+\alpha\Delta x$$ where $\alpha\to0$, if $\Delta x\to 0$. Now this is where my confusion starts: "The first summand on the right side of this equality depends on $\Delta x$ in a very simple way, namely it is proportional to $\Delta x$. It is called the differential of a function, at the point $x$ denoted by $dy=f'(x)\Delta x$."
But after in the end of the section, it was written that it may also be written as $$dy=f'(x)dx$$ But how?? $d$(infinitesimal) doesn't mean $\Delta$(change) right?
And then I came to "Differentiation of implicit functions"
"Suppose we find the derivative of $y$, where $y$ is a function of $x$ defined implicitly by $$F(x,y)=0$$ We give $x$ and $y$ the increments $\Delta x$ and $\Delta y$ respectively, such that they also satisfy the above relation. Thus, $$F(x+\Delta x,y+\Delta y)-F(x,y)=\frac{\partial F}{\partial x}\Delta x+\frac{\partial F}{\partial y}\Delta y+\alpha\sqrt{\Delta x^2+\Delta y^2}=0$$" But then they deduce that $$\frac{dy}{dx}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$$ But how did the $\alpha$ term vanish? Please explain I am new to this topic.

Answer 1

Your first question is legit. How we pass from $df = f'(x) \Delta x$ to $df = f'(x)dx$?

Well, let's calculate the derivative of the function $g(x) = x$, that we could call simply "the function $x$": $$\lim_{\Delta x \to 0} \frac{g(x + \Delta x) -g(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{x + \Delta x -x}{\Delta x} = \lim_{\Delta x \to 0} \frac{\Delta x}{\Delta x} = 1$$ So we can write: $$dg = g'(x)\Delta x = 1\cdot\Delta x = \Delta x$$ We can denote the function $g$ simply by $x$, and the previous formula becomes: $$dx = \Delta x$$

So we always can write $$df = f'(x)dx$$ instead of$$df = f'(x)\Delta x$$

Since the differential of the function $x$ is simple the increment.

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In order to be more precise we should write: $$df(x,\Delta x) = f'(x)\cdot \Delta x$$ i.e. the differential has two arguments, the point in with it is evaluated and the increment. Then, as we saw: $$dx(x,\Delta x) = \Delta x$$ In the second form we should write: $$df(x,\Delta x) = f'(x)\cdot dx(x,\Delta x)$$

So we write every differential as the product of the derivative and the "basic differential" $dx$.

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EDIT I add this little explanation on differential of linear functions:

If a function is linear, is in the form $l(x) = c \cdot x$ for a fixed number $c$. So: $$\lim_{\Delta x \to 0} \frac{l(x + \Delta x) - l(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{cx + c\Delta x - cx}{\Delta x} = c$$ and the differential: $dl(x,\Delta x) = c \Delta x$

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For the second question, you have to look at implicit function theorem

Answer 2

Since your question is long, I try to write something that might help you with thinking about the nature of $\alpha$.

Consider $g(h)=\frac{f(x+h)-f(x)}{h}$ for which we know that $\lim_{h\to0}g(h)=f'(x)$ (say all limits exist). Now lets write a Taylor's expansion for $g(h)$ at $0^+$:$$g(h)=f'(x)+hg'(0^+)+...$$ We have $$g'(h)=\frac{f'(x+h)h-f(x+h)+f(x)}{h^2}$$ hence \begin{align}\lim_{h\to 0}g'(h)&=\lim_{h\to 0}\frac{f'(x+h)h-f(x+h)+f(x)}{h^2}\\ &=\lim_{h\to 0}\frac{f''(x+h)h+f(x+h)-f'(x+h)}{2h}\\ &=\frac12\lim_{h\to 0}f''(x+h)=\frac12 f''(x)\\ \end{align}

Hence we have $$g(h)=f'(x)+h\frac12f''(x)+...$$ from here you could see that $$\frac{\Delta y}{\Delta x}=f'(x)+\Delta x \times \frac12 f''(x) + ...$$ or $$\Delta y=f'(x)\Delta x+\Delta x \times (\Delta x \frac12 f''(x) )+ ...$$i.e. $\alpha = \Delta x \frac12 f''(x)$. You could develop the same thing for bivariate or multivariate functions. knowing the nature of $\alpha$ in general helps understanding what happens to it.