Real function derivative
Definition
context | X,Y∈Jpo |
context | f:X→Y … continuous |
let | X∃≡{x∈X | limh→0f(x+h)−f(x)h∈Y} |
definiendum | f′:X∃→Y |
definiendum | f(x):=limh→0f(x+h)−f(x)h |
Discussion
To constructing f′ from f, we must restrict the domain X to X∃ 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.
Notation
f′(x)≡∂f(x)∂x≡∂f∂x |
---|
More generally we can write the differential with successive evaluation of the derivative at a point g(x) in the following ways:
∂f(x)∂x(g(x))≡∂f∂x(g(x))≡∂f(x)∂g(x)≡∂f∂g(x)≡∂∂y|y=g(x) f(y)≡(∂∂yf(y))y=g(x) |
---|
Note that ∂f(x)∂g(x) or ∂f∂g(x) might easily be misread: The expression ∂exp(x6)∂x2 is taken to be (∂∂yexp(y6))y=x2 and not (∂∂yexp(x3))y=x2. 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 ∂f(x1,x2)∂x2(g1(x1,x2),g2(x1,x2)). 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 ∂f∂y, denote ∂f(x,y)∂y(x,y).
Theorems
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
f'(x)=\dfrac{f(x+h)-f(x)}{h}\left|_{h=0}\right.
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:
f'(x)=\mathrm{lim}_{h{\to}0}\dfrac{f(x+h+h^2g(h))}{h}
- For a function H that eventually diverges monotonically, we have
f'(x)=\mathrm{lim}_{r\to\infty}\,H(r)\left(f(x+\frac{1}{H(r)})-f(x)\right)
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]
Reference
Wikipedia: Derivative of a function
Calculus developed with infinitesimals (non-standard analysis) instead of limits: http://www.math.wisc.edu/~keisler/calc.html