It is proved that the differentiable function must be continuous

Let y = f (x) be differentiable at point x, that is, Lim Δ Y / Δ x (Δ x approaches 0) = f ′ (x). From the relationship between the function with limit and infinitesimal, we know that Δ Y / Δ x = f ′ (x) + α, where α is the infinitesimal when Δ x approaches 0. By multiplying both sides of the above formula by Δ x, we can get Δ y = f ′ (x) Δ x + α Δ X. thus, when Δ x approaches 0, y approaches 0, The function y = f (x) is continuous at point x (according to the definition of function continuity), so it must be continuous

Double limit and quadratic limit of binary function
Is this proposition correct (better to have proof)
If one of two quadratic limits exists and the other does not exist for the same limit process, the double limit of the process does not exist

There is only one conclusion between the repeated limit and the double limit, that is, if they all exist, they must be equal, and nothing else can be deduced from each other

Let z = f (3x, X-Y), where f is a differentiable function, and find &; Z / &; X, &; Z / &; y

Let f (U, V, u = 3 x, u (U, V, u = 3 x, v = X-Y, then, let \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\; V) &; Z / &; y = (&; f

On the existence but non differentiability of partial derivatives of binary functions
The book says "for the function f (x, y) = {XY / radical (x ^ 2 + y ^ 2), x ^ 2 + y ^ 2 is not equal to 0
0 , x ^2+y^2=0
At point (0,0), there are FX (0,0) = 0 and FY (0,0) = 0, so“
I want to know how to get FX (0,0) = 0? The origin is separated from the surrounding definition. In this case, how to find the partial derivative?

For example, we can't use the derivative formula to find the derivative of the critical point of this piecewise function. We should use the definition of derivative to find it. According to the definition of partial derivative, in this problem, f'x (0,0) = Lim [f (x.0) - f (0,0)] / x = LIM (0-0) / x = 0, so is the partial derivative of Y. you can find it yourself

When y is greater than 0 and less than the square of X, f (x, y) is equal to 1. When y is other points, f (x, y) is equal to 0. Try to discuss it in (0,0)
When y is greater than 0 and less than the square of X, f (x, y) is equal to 1. When y is other points, f (x, y) is equal to 0. Try to discuss its limit at (0,0)

Unlimited
Follow the line y = 0.5x ^ 2
The limit is 1
Go along the y-axis
The limit is 0
complete
And all the points on the x-axis and on the y-axis

Finding dy / DX of chain rule
y＝√x＋1／x

Dy / DX = 1 / 2 √ X-1 / x ^ 2 doesn't need chain rule

For example, the derivative of a function is not necessarily differentiable
F (x) is continuous in the open interval (a, b) and has the first derivative, but for some point x0 ∈ (a, b), f '(x0) does not exist
Give me an example of F (x) satisfying the above conditions
Sorry, the wrong number. It should be f '' (x0) does not exist

If x is on (- infinity, 0), f (x) = - X
If x is on (0, + infinity), f (x) = X
The above functions are continuous in the domain of definition and continuous at x = 0, but the left limit is not equal to the right limit, i.e. f '(x0) does not exist

Who can give an example that is not a piecewise function to show that the original function is differentiable but its derivative is not necessarily continuous

X * e ^ x is the power of X multiplied by E; when x ≠ 0
f(x)=
0 ；x=0
It's good to go upstairs~~

The derivative dy / DX of the implicit function y determined by the equation E ^ y · X-10 + Y & # 178; = 0

The derivation of X on both sides is as follows:
(e^y)'·x+e^y·(x)'+2y·y'=0
e^y ·y'·x+e^y+2y·y'=0
y'(x·e^y+2y)=-e^y
dy/dx=y'=-e^y/(x·e^y+2y)

Let z = Z (x, y) be determined by the equation Z + x = e ^ (Z-Y), and find the partial derivative δ ^ 2 Z / δ y δ X

your answer here

It is proved that the differentiable function must be continuous