Real function derivative


context $X,Y\in \mathfrak J_o^p$
context $f:X\to Y$ … continuous
let $X_\exists\equiv\{x\in X\ |\ \mathrm{lim}_{h\to 0}\frac{f(x+h)-f(x)}{h}\in Y\}$
definiendum $f':X_\mathrm{\exists}\to Y$
definiendum $f(x) := \mathrm{lim}_{h\to 0}\dfrac{f(x+h)-f(x)}{h} $


To constructing $f'$ from $f$, we must restrict the domain $X$ to $X_\exists$ where, per definition, the required limit exists. We could alternatively omit this and define $f'$ to be a partial function.

In this entry we defined how to pass from a given function $f$ to the derivative $f'$. This process is in fact functional in $f$ and can hence be internalized, see Fréchet derivative.


$f'(x) \equiv \frac{\partial f(x)}{\partial x} \equiv \frac{\partial f}{\partial x} $

More generally we can write the differential with successive evaluation of the derivative at a point $g(x)$ in the following ways:

$\frac{\partial f(x)}{\partial x}(g(x)) \equiv\frac{\partial f}{\partial x}(g(x))\equiv\frac{\partial f(x)}{\partial g(x)} \equiv\frac{\partial f}{\partial g(x)} \equiv \frac{\partial}{\partial y}|_{y=g(x)}\ f(y) \equiv\left(\frac{\partial}{\partial y}f(y)\right)_{y=g(x)}$

Note that $\frac{\partial f(x)}{\partial g(x)}$ or $\frac{\partial f}{\partial g(x)}$ might easily be misread: The expression $\frac{\partial \mathrm{exp}(x^6)}{\partial x^2}$ is taken to be $\left(\frac{\partial}{\partial y}\mathrm{exp}(y^6)\right)_{y=x^2}$ and not $\left(\frac{\partial}{\partial y}\mathrm{exp}(x^3)\right)_{y=x^2}$. I.e. functions are always derived w.r.t. their proper arguments alone.

If the domain of $f$ is higher dimensional, some of the above notations don't work anymore. We must e.g. use the unambigous notation $\frac{\partial f(x^1,x^2)}{\partial x^2}(g_1(x^1,x^2),g_2(x^1,x^2))$. Since this is cumbersome, the variable names are usually implicitly understood to be held fixed, e.g. in writing “$f(x,y)$” once and then have $\frac{\partial f}{\partial y}$, denote $\frac{\partial f(x,y)}{\partial y}(x,y)$.


$f^{(n)}(x) = \dfrac{1}{h^n}\lim_{h \to 0}\sum_{0 \le m \le n}(-1)^m {n \choose m}f(x+(n-m)h)$

This can simply be extended to non-natual $n$ and is then called Grünwald–Letnikov derivative.

For a real function $\mathbb R\to\mathbb R$ to have a derivative at a point, it must be continuous at that point. (This is wrong for the derivative in higher dimensions, e.g. consider the function $f(x,y):=\frac{xy}{x^2+y^2}$ on $x,y\neq 0$ and else $f(0,0):=0$.)

Fermats observation

For polynomials (and other nice functions), we have that the value of the derivative can be obtained algebraically as follows


E.g. $p(x) := 2x^2-x+3 \implies p(x+h)-p(x) = 2(2xh+h^2)-h = h\,((4x-1)+2h)$

It also works for $f(x):=\frac{1}{a+bx}$, but not for many variations of that.

Other formulations
  • If $g$ has a power series expansion, then also an additional term $h^2g(h)$ in the argument can't matter:


  • For a function $H$ that eventually diverges monotonically, we have


H[r] = 3 + Log[r]; (* axillary growing function *)

d[f_][x_] = Limit[H[r] (f[x + 1/H[r]] - f[x]), r -> Infinity];

d[#^n &][x]


Wikipedia: Derivative of a function

Calculus developed with infinitesimals (non-standard analysis) instead of limits:


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