Why can the second derivative f'' be written as d²f/dx²? #Science##Math#For this problem, we need to start from the beginning: ———— for any real function f(x

Why can the second derivative f'' be written as d²f/dx²?

#Science ##Math#

On this subject, we need to start from the beginning:

————

For any real function f(x), we denote the real number (axis) as R, then x is the variable in R, and let X be a point in R, and h is the amount of change of x near the X point, i.e.,

x=X+h

To say that f(x) is differentiable at point X is to refer to the amount of change in f(x) near point X:

Δfx(h)=f(x)-f(X)=f(X+h)- f(X)

Approximate a straight line that passes through (X, f(X)):

lx(h)=Kxh

where Kx is the slope of the straight line lx.

The more precise meaning of "approximation" here is:

When h→0, Δfx(h)→lx(h), which is Δfx(h)/lx(h)→1

So there is (the subscript of the following lim is h→0),

1=lim Δfx(h)/lx(h)=lim (f(X+h)- f(X))/(Kxh)=1/Kxlim (f(X+h)- f(X))/h

And so you get,

Kx=lim (f(X+h)- f(X))/h

We say that Kx is the derivative of f at the X point, denoted f'x=Kx, and that the line lx is the differentiation of f at the X point, and that dfx=lx, so that there is,

dfx(h)=lx(h)=Kxh=f'xh

If f(x) is differentiable at every point X in R, then X can be considered a variable, and the derivative f'x is a unary function of X (called a derivative), and the differential dfx(h) is a binary function of X and h. According to the writing habit, we move X from the subscript to () and replace it with x, so that we get,

df(x,h)=f'(x)h (1)

thereinto

f'(x)=lim (f(x+h)- f(x))/h

For the identity function I(x)=x yes,

I'(x)=lim ((x+h)- x)/h=1

Further, the differentiation is simultaneously on both sides of x=I(x), and there is,

dx=dI(x, h)=I'(x)h=1h=h=I(h)

> Geometrically, the tangent at any point of a line is itself.

Thus (1) becomes,

df(x, h)＝f'(x)dx

Since the right side of the equation dx has hidden h, the left side does not explicitly declare h, so the equation is rewritten as:

df(x)＝f'(x)dx ①'

And then there are,

f'(x)＝df(x)/dx

Thus we get a differential representation of the derivative:

df/dx＝f'

Then differential df(x), according to (1)' and dx=h, yes,

d(df)(x)＝(df(x))'dx＝(f'(x)dx)'dx＝(f'(x)h)'dx＝(f'(x))'hdx＝f''(x)dxdx

>note: 'Is to derive x, h is considered a constant.'

So there is,

f''(x)＝d(df)(x)/dxdx

remember

d²f＝ddf＝d(df)，dx²＝(dx)²＝dxdx

This gives the differential form of the second derivative:

d²f/dx²=f''=d(df/dx)/dx

Finally, by iterating on, we can derive differential forms of derivatives of any n (natural number) order:

dⁿf/dxⁿ＝f⁽ⁿ⁾

>规定:f⁽⁰⁾=f,d⁰f=f,dx⁰=1。