Why is the derivative of LN x 1 / x? Please write the derivation

Why is the derivative of LN x 1 / x? Please write the derivation

The derivative of F (x) = limx1 - > 0 [f (x + x1) - f (x)] / X1 = limx1 - > 0 [ln (x + x1) - LNX] / X1 = limx1 - > 0 [ln (1 + X1 / x)] / X1 = limx1 - > 0 1 / X * x / X1 * ln (1 + X1 / x = 1 / X * limx1 - > 0 ln (1 + X1 / x) ^ X / X1 = 1 / X * lne = 1 / X

1. Let f (x) = 5 (x ^ 10) * (e ^ 20) find f (20) (x), 2. Let y = ln / [(1 + SiNx) / (1-sinx)], find (d ^ 2) y / D (x ^ 2), that is, find the second derivative of the function

With MATLAN software, I didn't download it. You can enter it and press enter

How to find the derivative of 1-sinx / 1 + SiNx under y = LN

Geometric meaning of second order mixed partial derivatives?

What the first floor said is the geometric meaning of the first partial derivative
Second order mixed partial derivative has no geometric meaning
Of course, it must be done, not impossible
F〃xy（x0,y0）＝（F′x（x0,y）'y（y0）
In other words, let's first make a unary function Φ (y) = f ′ x (x0, y), and the slope of the tangent of the image z = Φ (y) at (Y0, Φ (Y0)) is the "geometric meaning" of F ″ XY (x0, Y0)
So, let's just say that the second mixed partial derivative has no obvious geometric meaning

What is the meaning of mixed partial derivatives?
For example, geometry

In multivariate function, if the variables of several partial derivatives are not the same, it is called mixed partial derivative. For example, the partial derivative of X is obtained first, and then the partial derivative of Y is obtained

Second order mixed partial derivatives
If the second-order mixed partial derivative is continuous at the stagnation point of the binary function, is the second-order mixed partial derivative equal to zero? That is, if the first-order partial derivative of the function f (x, y) to X and Y is equal to zero at the given point P (x0, Y0), and there is a continuous second-order mixed partial derivative of F at the other point P, is the second-order mixed partial derivative equal to zero?

It is not necessary that the stationary point is the point where the first partial derivative of X and Y is equal to 0. Whether the extreme value is obtained at this point is given by the positive and negative of AC-B ^ 2, a = FXX, B = fxy, C = FYy

On the problem of finding the second order partial derivative,

1.
az/ax
=f1*(2x)+f2*(1/y)
two
First, find the first partial derivative
az/ax=f1*(2x)+f2*(1/y)
The second partial derivative is obtained
a^2z/axy
=f12*(-x/y^2)*(2x)+f22*(-x/y^2)*(1/y)+f2*(-1/y^2)
=f12*(-2x^2/y^2)+f22*(-x/y^3)+f2*(-1/y^2)
Only use the formula:
If z = f (a (x, y), B (x, y))
Then: AZ / AX = F1 * a '+ F2 * B'
Where, F 1 and F 2 are z = f (a (x, y), B (x, y)) for partial derivatives of a and B, and a ', B' for partial derivatives of a and B for partial derivatives of X
And AZ / ay is the same
If you don't understand, please ask

Second order partial derivatives

As shown in the picture

The second order partial derivative problem
Given 2x + 2Y + Z + 8xz + 8 = 0, where x and y are functions of Z, find the second order and mixed partial derivatives of Z to X and y

2X^2 + 2Y^2 +Z^2 +8XZ +8=0
The partial derivative of the above formula for X is: 4x + 2Z * Z '(partial derivative of x) + 8Z + 8xz' (partial derivative of x) = 0
We can get Z '(partial derivative of x), and the second derivative is similar to this

The problem of solving mixed partial derivatives of second order
What is the formula d = FXX (x, y) FYy (x, y) - [fxy (x, y)] ^ 2? Is it used to judge whether a point is a local extreme value?

It's just a decision formula. The second problem, yes,