Who can have partial derivative! Z = 1 / XY, find the partial derivative of (x2 + Y2 + Z2) under α Z / α X. r = radical If all elements in a row of a determinant are zero, why is the determinant Let z = 1 / XY, then α X / α y =? Let (x1, X2,...) X3) is a sample taken from population x, then the sample variance s =? Find the partial derivative under r = root sign (X & # 178; + Y & # 178; + Z & # 178;)

Who can have partial derivative! Z = 1 / XY, find the partial derivative of (x2 + Y2 + Z2) under α Z / α X. r = radical If all elements in a row of a determinant are zero, why is the determinant Let z = 1 / XY, then α X / α y =? Let (x1, X2,...) X3) is a sample taken from population x, then the sample variance s =? Find the partial derivative under r = root sign (X & # 178; + Y & # 178; + Z & # 178;)

When seeking partial derivatives, for whom, the others can be regarded as constants

If the directional derivatives of F (x, y) exist in any direction at (x0, Y0), then f (x, y) has partial derivatives at (x0, Y0)
What's wrong with that?

1. Since there are directional derivatives in any direction, naturally there are directional derivatives in the X direction, and the directional derivatives in the X direction are partial derivatives in the X direction

Finding the partial derivatives AZ / ax... And a ^ 2Z / ax ^ 2 of the function z = ln (x ^ 2 + y ^ 2)

az/ax=2x/（x^2+y^2)
A ^ 2Z / ax ^ 2 = the square of 2 (- x ^ 2 + y ^ 2) / [(x ^ 2 + y ^ 2)]

Let f (x), G (x) be continuous on {a, B}, differentiable in (a, b), and f '(x) = g' (x), X ∈ (a, b). It is proved that there exists a constant C such that f (x) = G (x) + C
And X ∈ {a, B}
Urgent answer, thank you
Please prove it in detail

Let f (x) = f (x) - G (x)
Then f '(x) = 0
For any two points x, y in [a, b]
According to Lagrange's mean value theorem, it is easy to get f (x) = f (y)
Therefore, f (x) is constant and it is easy to draw a conclusion

Let f (x) be continuous on [a, b], differentiable in (a, b), and f (a) = f (b) = 0. It is proved that there exists C ∈ (a, b) such that f '(c) + F (c) = 0
Let f (x) be continuous on [a, b], differentiable in (a, b), and f (a) = f (b) = 0. It is proved that there exists C ∈ (a, b) such that f '(c) + F (c) = 0
To prove that f '(c) + [f (c)] ^ 2 = 0

Let f (x) = e ^ x * f (x)
Then f (a) = f (b) = 0
From the mean value theorem
There exists C ∈ (a, b), f '(c) = e ^ CF (c) + e ^ CF' (c) = e ^ C (f '(c) + F (c)) = 0
That is, f '(c) + F (c) = 0

How to prove that the partial derivative of this function is discontinuous~~
 

Teacher, please help me to look at this problem again and prove that the function f (x, y) = √ x ^ 2 + y ^ 2 is continuous at (0,0), but there is no partial derivative at (0,0)

You see that: partial derivative must be continuous, continuous not partial derivative. Examples of theorems in the book

F (x) = (x ^ 1 / 2) / [(1 + x) ^ 1 / 2 + 1] it is proved that at x = 0, the right derivative is right continuous, and there is no process to find the right derivative

F (x) = √ X / [(1 + x) ^ 1 / 2 + 1], f (0) = 01. Lim [f (x), X - > 0 +] = 0 = f (0), that is, f (x) is right continuous at x = 0. F '+ (0) = Lim [[f (x) - f (0)] / (x-0), X - > 0 +] = Lim [[(1 + x) ^ 1 / 2 + 1] / √ x, X - > 0 +] = ∞ that is, the right derivative of F (x) does not exist at x = 0

The polynomial x ∧ 4-3x & sup3; + 3x & sup2; + MX + n can be divisible by X & sup2; - 3x + 2 to find the value of M.N

x²－3x＋2
=（x-1)(x-2)
Then x = 1, x = 2
x∧4－3x³＋3x²＋mx＋n=0
∴ 1-3+3+m+n=0
m+n=-1 （1）
16-24+12+2m+n=0
2m+n=-4 （2）
(2) - (1) get
m=-3
n=2

If 2x ＾ 3 + X ＾ 2 + kx-2 can be divided by 2x + 1 / 2, the value of K is obtained

The division of polynomials. X ^ 2 + 0.25x - 4_______________________________ 2x+1/2 / 2x^3 + x^2 + kx - 2 .√ 2x^3 + 0.5x^2 .￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣ .0.5x^2 + kx .0.5x^2 + 0.125x .￣￣￣￣￣￣￣￣￣...