Two partial derivatives of a binary function exist at a certain point. Can we deduce that it is defined in a neighborhood of the change point?

What we should study in this problem is the definition of partial derivative. Only if there is a definition in the critical field of a point, we can get the partial derivative. On the contrary, we can get the definition in the critical field of that point. Refer to the definition of partial derivative

The function z = f (x, y) has continuous partial derivatives at P (x0, Y0). It is proved that the direction of P 0 along the tangent direction of the isoline
The function z = f (x, y) has continuous partial derivatives at P0 (x0, Y0). It is proved that the directional derivative along the tangent direction of the isoline at P0 is zero

Let f (x0, Y0) = C, then the isoline equation of Z = f (x, y) at point P0 is f (x, y) = C, where y is understood as the function of X, and the derivative rule of composite function is used to get f'x + f'y * dy / DX = 0, that is, the tangent slope of isoline at point P0 Tan θ = dy / DX = - f'x / f'y, so sin θ = - f'x, cos θ = f'y, so the directional derivative along the tangent direction of isoline at point P0 = f'x * cos θ + f'y * sin θ = - f'xf'y + f'xf'y = 0

If the binary function z = arctgxy, then the value of the partial derivative of Z (x, y) with respect to X at point (1,1) is ()

Partial Z / partial x = Y / [1 + (XY) ^ 2]
Then x = 1, y = 1
Partial Z / partial x = 1 / [1 + (1 × 1) ^ 2] = 1 / 2

The third power of X + m times the second power of X + 2x + 1 can be divided by (x + 1). How much is the value of X

The cubic power of X + m times the quadratic power of X + 2x + 1 can be divided by (x + 1), which shows that the previous formula = (x + 1) is cubic
One by one, we get x = 0

The quadratic trinomial x ^ 2 + 7x-m of X can be decomposed into (x + 3) (x-m), then the values of M and N are?

Wrong title. Which is n?
If m in quadratic trinomial x ^ 2 + 7x-m is n, M = - 4, n = - 12
If (x + 3) (x-m), where m is n, M = - 12, n = - 4

On the quadratic trinomial X & # 178; + 7x-m of X can be decomposed into (x + 3) (x-n), then the values of M and N are?

(x+3)(x-n)=x^2+(3-n)x-3n=x^2+7x-m 3-n=7 n=-4 m=3n=-12

The polynomial 7x ^ m + (k-1) x ^ 2 - (2n + 4) X-6 is a cubic trinomial about X, and the coefficient of the quadratic term is 1
Elder brother and elder sister, please help

m=3,k-1=1,2n+4=0
So m = 3, k = 2, n = - 2
m+n-k=-1

Quadratic trinomial x ^ 2 + 7x + 12 = (x + 3) (x +)

x^2+7x+12=(x+3)(x+4).

Point C is the point on line AB, m and N are the key points of line AB and AC respectively. If AC = 9 cm, find the length of line Mn

Let AB length be l > 9, then a (0) B (L) m (L / 2)
The length of AC is 9 C (9) n (4.5)
The length of Mn is 1 / 2-4.5
That is Mn = AB / 2-4.5

It is known that there are two points m and N on the line ab. point m divides AB into two parts 2:3 and point n divides AB into two parts 4:1. If Mn = 3cm, the length of AM and Nb can be obtained

∵ point m divides AB into 2:3 parts, ∵ am = 25ab; ∵ point n divides AB into 4:1 parts, ∵ an-am = Mn, ∵ 45ab-25ab = 3, ∵ AB = 7.5cm, ∵ am = 25ab = 3cm. ∵ NB = ab-an = 15ab = 1.5cm. So am = 3cm, Nb = 1.5cm

Two partial derivatives of a binary function exist at a certain point. Can we deduce that it is defined in a neighborhood of the change point?