Prove that the function of two variables is differentiable Let Lim [f (x, y) - f (0,0) + 2x-y] / √ x ^ 2 + y ^ 2 = 0, prove that f (x, y) is differentiable at point (0,0) (x,y)→(0,0) He said: F (x, y) - f (0,0) + 2x-y = O (ρ), (when (x, y) → (0,0)), we can get that f (x, y) is differentiable at point (0,0). How can we get differentiable?

Prove that the function of two variables is differentiable Let Lim [f (x, y) - f (0,0) + 2x-y] / √ x ^ 2 + y ^ 2 = 0, prove that f (x, y) is differentiable at point (0,0) (x,y)→(0,0) He said: F (x, y) - f (0,0) + 2x-y = O (ρ), (when (x, y) → (0,0)), we can get that f (x, y) is differentiable at point (0,0). How can we get differentiable?

The definition of the function z = f (x, y) the total increment of the point (x, y) at the point (x, y) the definition of the function z = f (x, y) the full increment of the function z = f (x, y) the full increment of the function z = f (x, x, y + Δ x, y + Δ y) - f (0,0), the full increment of the function z = f (x = x = 0, then the full increment of the function z = f (x = f (x, x, x, Δ Z = f, Δ z = f, Δ x, Δ x, Δ y is replaced by X, y to express the symbol Δ x, Δ x, Δ x, Δ x, the problem (x, y) → (0,0,0,0,0) when (x (x, y) → (0,0,0,0,0,0) when the function f (x (x (x, x, x, y) → (0,0,it can be divided into two parts

Geometric meaning of multivariate function gradient?

If the multivariate function is regarded as height, its gradient is the direction of the steepest ascent
If the multivariate function is regarded as potential energy, the negative value of its gradient is the local force

What is the geometric meaning of the "value" of the modulus of gradient?
Of course, he represents the maximum value of the directional derivative of the function. Does his personal feeling (for example) represent the "slope" of the slope in the steepest direction at a certain point,
are you sure?

Your understanding is correct:
The "slope" of a hillside in the steepest direction at a point

It is known that the triangle ABC is an equilateral triangle, extending BC to D, extending Ba to e, and making AE = BD, connecting CD and de
Find multiple solutions

No, there is a solution to this problem! I think it's easier to use vector to solve this problem. To prove EC = ed, we only need to prove (bc-be) * (bc-be) = (bd-be) square, that is, BC square - 2 * BC · be + BD square = BD square - 2 * BD · be + be square

Given that f (x) = logax (a is greater than 0, and a is not equal to 1), if 2, f (x1), f (x2), f (x3),, f (xn), 2n + 4, find the common term of {an} in an arithmetic sequence

2, f (x1), f (x2), f (x3), f (xn), 2n + 4, in arithmetic sequence (n + 2 items in total)
If the tolerance is D, then 2n + 4 = 2 + (n + 2-1) d
d=2
So f (xn) = 2 + nd = 2n + 2
That is: loga (an) = 2n + 2
an=a^(2n+2)

How to prove that four points are coplanar with space vector?

Arbitrary connection, you can get three vectors, and then the determinant is zero, you can get coplanar

It is known that: as shown in the figure, in △ ABC, ab = AC, AE is the angular bisector, BM bisector ∠ ABC intersects AE at point m, through ⊙ o of B and M intersects BC at point G, intersects AB at point F, FB is exactly the diameter of ⊙ O. (1) prove that AE is tangent to ⊙ o; (2) when BC = 4, AC = 6, find the radius of ⊙ o

(1) It is proved that: if OM is connected, then ∠ OMB = ∠ OBM = ∠ MBE and ⊙ AB = AC, AE is an angular bisector, ⊥ BC, ∵ OMB + ∠ BME = ∠ MBE + ∠ BME = 90 degree, ∵ amo = 90 degree, ⊙ AE is tangent to ⊙ O. (2) by AE and ⊥ o, AE ⊥ BC ∥ om ∥ BC ∥ AOM ∥ Abe ∥ ombe = aoab ∥ BC = 4 ∥ be = 2, ab = 6, that is, R2 = 6 − R6, r = 32

The monkey king led a group of monkeys to pick peaches. After finishing work in the afternoon, the monkey king began to distribute. If the big monkey was divided into five, the monkey king could keep 10. If the big monkey and the small monkey were divided into four, the monkey king could keep 20. In this group of monkeys, how many more were the big monkey (excluding the monkey king) than the small monkey?

The topic is not complete, lack a condition, it should be like this: Monkey King leads a group of monkeys to pick peaches. After finishing work in the afternoon, Monkey King begins to distribute. If the monkey is divided into 5, the monkey is divided into 3, Monkey King can keep 10. If the monkey is divided into 4, Monkey King can keep 20. In this group of monkeys, Monkey King (excluding Monkey King) is more than monkey_______ ...

The unit vector parallel to vector a = (12,5) is ()
A. (1213, − 513) B. (− 1213, − 513) C. (1213513) or (− 1213, − 513) d. (− 1213513) or (1213, − 513)

Let the unit vector b = (x, y), | a | = 13 which is parallel to the vector a = (12, 5), so a = ± 13bb = (1213513), or B = (− 1213, − 513), so C

(1) As shown in Fig. 1, the bisector AE of ∠ bad intersects with the bisector ce of ∠ BCD at point E, ab ‖ CD, ∠ ADC = 40 ° and ∠ ABC = 30 ° to calculate the size of ∠ AEC; (2) as shown in Fig. 2, the bisector AE of ∠ bad intersects with the bisector ce of ∠ BCD at point E, ∠ ADC = m ° and ∠ ABC = n ° to calculate the size of ∠ AEC; (3) as shown in Fig. 3, the bisector AE of ∠ bad intersects with the bisector ce of ∠ BCD at point E, then the bisector AE and ∠ ADC Is there still some equivalent relationship between ABC and ABC? If there is, please write your conclusion and give proof; if not, please give reasons

(1) In the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, and in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, and in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, in the end, the end, in the end, in the end, the end, in the end, the× (40 ° + 30 °) = 35 °; (2) ∵ CE bisection ∠ BCD, AE As a result, it is found that the "bad" in the end of the paper that the PointF, ∵ ∠ BFD = ∠ B + ∠ bad, ∵ ∠ BCD = ∠ BFD + ∠ d = ∠ B +In this paper, the author introduces the basic principle of AEC, which is as follows: bad + bad, ∵ CE bisection ∵ BCD, AE bisection ∵ bad ∵ ECD = ∵ ECB = 12 ∵ BCD, ∵ ead = ∵ EAB, ∵ e = ∵ B + ∵ EAB - ∵ ECB = ∵ B + ∵ bae-12 ∵ BCD = ∵ B + ∵ bae-12 (∵ B + ∵ bad + ∵ d) = 12 (