How can a discontinuous function have partial derivatives? Functions of one variable are like this, so functions of many variables are different?

How can a discontinuous function have partial derivatives? Functions of one variable are like this, so functions of many variables are different?

Although the function f (x, y) is discontinuous at a certain point (x0, Y0), f (x, Y0) or F (x0, y) may be continuous, that is, f (x, Y0) is differentiable to X or F (x0, y) is differentiable to y, that is, the partial derivative of F (x, y) exists

Finding the second order partial derivative of bivariate function z = x ^ 2ye ^ y

As a result, we are going to be the \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\; y = 2xe ^ y + 2xye ^ y &; &; Z / &

The second order partial derivative of binary function, the problem of continuity
In the following binary functions, f '' (XY) (0,0) is not equal to f '' (YX) (0,0)
F (x, y) = (x * y ^ 2) * (x ^ 2-y ^ 2) / (x ^ 2 + y ^ 2), if (x, y) is not equal to (0,0), if (x, y) = (0,0), f (x, y) = 0
In order to solve the problem of whether the continuous partial derivatives are equal or not, we need to set y = KX and use the substitution method to carry out the troublesome calculation,

One is to accumulate and remember some examples, including conclusions and methods
Another is to accumulate some experience on how to choose methods. Knowledge includes the relationship of limit, continuity, partial derivation, differentiability, etc., which varies from problem to problem. For example, when it is necessary to judge differentiability, if there is no limit or discontinuity, then it is not differentiable

A binary function for the partial derivative of the problem, in fact, relatively simple
Find the partial derivatives of the function f (x, y) = x ^ + 2XY + y ^ at points (1,3) for X and y
Taking y as a constant, the derivative of function f (x, y) to X is obtained
f(x,y)=2x+2y
Taking x as a constant, the derivative of function f (x, y) to y is obtained
f(x,y)=2x-2y
Substituting points (1,3) into the above two formulas, we obtain
Omit
In fact, the answer is not complete, I understand the omitted part, but my problem is that the previous "function f (x, y) to x derivative, get: F (x, y) = 2x + 2Y" do not know how the "2x + 2Y" is calculated. I hope to write the steps and instructions. 80 points
>>>
F (x, y) = 2x + 2Y note: there is an X after F
F (x, y) = 2x-2y note: there is a y after F
^Representative number: 2
I wrote it in C language

The solution of partial derivative is to treat other variables as constants when you calculate partial derivative of a variable. The partial derivative of x ^ + 2XY + y ^ to X is the sum of partial derivatives of three parts to X. where x ^ is 2x, 2XY can be regarded as 2Y * x, so the partial derivative is 2Y (y is a constant), and Y ^ is a constant, so the partial derivative to X is naturally 0
The partial derivatives of Y are the same

If a partial derivative of a function of two variables is discontinuous at a point, is the function nondifferentiable at that point? How to prove nondifferentiability

If a partial derivative of a function of two variables is discontinuous at a point, is the function nondifferentiable at that point?
not always.
If we want to prove that it is not differentiable, how should we prove it
First, let's see if there is a partial derivative
If it doesn't exist, then it's not differentiable
If there is, then
And then prove it
Is the (Δ z-dz) / ρ limit 0
If it is 0, then it is differentiable, otherwise it is not differentiable

Let f (x) have continuous derivatives and satisfy the equation, ∫ (0-x) (x-t + 1) f '(T) DT = x ^ 2 + e ^ x-f (x), find f (x)

Let x = 0:0 = 1-f (0), f (0) = 1, left = x ∫ (0 → x) f '(T) DT - ∫ (0 → x) (t-1) f' (T) DT = x (f (x) - f (0)) - ∫ (0 → x) (t-1) f '(T) DT = XF (x) - X - ∫ (0 → x) (t-1) f' (T) DT: F (x) + XF '(x) - 1 - (x-1) f' (x) = 2x + e ^ x-f '(x) f (x) + XF' (x) - 1-xf '(x) + F' (T) DT

The upper and lower limits of the integral are [0, q], the integrand is XF (x), and f (x) is the derivative of F (x), so what is the integral equal to?

Partial integral method (integral limit is omitted): ∫ XF (x) DX = ∫ XDF (x) = QF (q) -∫ f (x) DX, the latter integral cannot be solved due to insufficient conditions

Let f (x) be differentiable in (0, + ∞), its inverse function be g (x), and ∫ [upper and lower limits (1, f (x))] g (T) DT = 1 / 3 * {x ^ (3 / 2) - 8}, then f (x) can be obtained
I already know that the original function is f (x) = root x + C
The answer is C = - 1
It is suggested that f (4) = 1

Let f (x) = 1, then ∫ [upper and lower limits (1,1)] g (T) DT = 1 / 3 * {x ^ (3 / 2) - 8}, left is 0,
Let 1 / 3 * {x ^ (3 / 2) - 8} = 0, so when x = 4, the above formula is 0,
F (4) = 1

F (x) = ∫ f (T / 2) DT the derivative of integral sign below 0 and above 2x f (x) is

Let the original function of F (x) be f (x), that is, f (x) = F & # 39; (x), then

The derivative of differential is integral, the original function of integral is differential. Is this sentence probably right? Is it wrong

It's not right at all. The derivative of differential and the original function of integral are twice differential or integral
To integrate a function and to differentiate it, these two operations are reciprocal operations
The process of finding the original function is indefinite integral operation; the process of finding the derivative is differential operation
The differential of a function is slightly different from its derivative. The differential is the linear increment (change) of a function, while the derivative is the rate of change (change of function value / change of independent variable)