# What is the mathematical truth behind the Leibniz notation in differentiating twice or more?

So $f: \mathbb{R} \to \mathbb{R}$ is $n>1$ (or more) times differentiable.

The notation of the first derivative makes perfect "sense" with regard to what's going on:

$$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \equiv \frac{df}{dx}$$

The second makes me tilt my head a bit (to no effect):

$$\lim_{h \to 0} \frac{\frac{df}{dx}\big|_{x+h} - \frac{df}{dx}\big|_{x}}{h} = \frac{d}{dx} \frac{df}{dx} = \frac{d^2 f}{dx^2}$$

This notation looks like as if:

$$\frac{d^2 f}{dx^2} = \lim_{h \to 0} \frac{(f(x+h)-f(x))^2}{(x+h)^2-x^2}$$ But I couldn't find any sense in that..

Now, I was told that this notation has complete (more advanced) mathematical sense.

I'd like to know where to look for it.

• I'm starting to have some ideas just by asking this out loud.. – user76568 Sep 11 '14 at 16:55
• It is quite common that ironically you find your answer to your question when you ask it. But honestly, I do not see what you are asking here. Are you asking whether the formula that follows "as if" is correct or not? – Lord Soth Sep 11 '14 at 17:10
• Not exactly. I'm asking for a mathematical analysis (Or a reference to one) that eliminates the (my) question mark that is hovering above this notation. Regarding the as if formula; I did not ask about It's validity. It's just (one of) the odd things one can try out. maybe expanding something in it and neglecting things which are too tiny, tho. I'll try – user76568 Sep 11 '14 at 17:16
• Your interpretation reads the formula as $(df)^2/d(x^2)$, which is not at all the same as $d^2 f/(dx)^2$. – Hans Lundmark Sep 11 '14 at 17:41
• @HansLundmark I thought the $d$ is essentially a function taking $f$ and $x$ or $x^2$ as the arguments (Like $\sin^{2}(x) = (\sin (x))^2$) and that's why I thought/think that $d^{2}f = (df)^2$ and $dx^{2}=d(x^2)$. Is this wrong for a fact? – user76568 Sep 11 '14 at 17:50

Assuming that $f(x)$ is twice differentiable, we have $$\frac{d^2}{dx^2}[f(x)]=\frac{d}{dx}\left[\frac{d}{dx}[f(x)]\right]$$ $$=\frac{d}{dx}\left[\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}\right]$$ $$=\lim\limits_{h\to 0} \frac{\frac{d}{dx}[f(x+h)-f(x)]}{h}$$ $$=\lim\limits_{h\to 0} \frac{\lim\limits_{k\to 0}\frac{f(x+h+k)-f(x+k)-f(x+h)+f(x)}{k}}{h}$$ $$=\lim\limits_{h\to 0}\lim\limits_{k\to 0} \frac{f(x+h+k)-f(x+k)-f(x+h)+f(x)}{hk}$$ You might come across the following formula for the $n^{\mathrm{th}}$ derivative $$\frac{d^n}{dx^n}[f(x)]=\lim_{h\to 0} \frac{1}{h^n}\sum_{k=0}^n (-1)^n\left(\begin{array}{c}n \\ k \end{array}\right)f(x+kh)$$ Although in some cases this limit does output a value that is equivalent to the $n^{\mathrm{th}}$ derivative, this formula is ultimately ill defined. A more precise definition would have $n$ limits as opposed to just one. Furthermore, there are cases in which this limit exists, but $f(x)$ is not $n$ times differentiable.
• @DanielV and k170: Can you guys give an example of a $n$ times ($2$ just for the example is enough for me) differentiable function that would render the formula in this answer erroneous? – user76568 Sep 12 '14 at 18:14
• @k170 $\lim_{h \to 0} \lim_{k \to 0} f(h, k)$ is the same as $\lim_{(h, k) \to (0, 0)} f(h, k)$ when they are defined, but the former is defined for more functions than the latter, as the latter requires that all paths to $(0, 0)$ reach the same limit. – DanielV Jul 24 '15 at 4:37