The U / V power of Z = UE u = x ^ 2 + y ^ 2 v = XY

The U / V power of Z = UE u = x ^ 2 + y ^ 2 v = XY

Z = UE ^ (U / V), where u = x ^ 2 + y ^ 2, v = XY,
z' = u'e^(u/v)+ue^(u/v)*(u'v-uv')/v^2
= e^[(x^2+y^2)/(xy)] {2x+(x^2+y^2)[2yx^2-(x^2+y^2)y/(xy)^2]}
= e^[(x^2+y^2)/(xy)] [2x+(x^2+y^2)(x^2-y^2)/(yx^2)]
= e^[(x^2+y^2)/(xy)] [2x+(x^4-y^4)/(yx^2)].
By rotation, we get
z' = e^[(x^2+y^2)/(xy)] [2y+(y^4-x^4)/(xy^2)].

Finding the total differential of Z: z = arcsin (XY) z = xsin (x + y)

Z=arcsin(xy)Z'xy=1/√(1-(xy)^2)(xy)'x=y(xy)'y=xdZ= Z'xy *(xy)'x dx+Z'xy*(xy)'y dy=ydx/√(1-(xy)^2 +xdy/√(1-(xy)^2Z=xsin(x+y)Z'x=sin(x+y)+xcos(x+y)Z'y=xcos(x+y)dZ=Z'x*dx +Z'y*dy=(sin(x+y)+xcos(x+y)) *...

If f (x) satisfies f (x + y) = f (x) + F (y) + 2XY, f (1) = 2, then f (- 3) =?
Solution: let x = 1, y = 0, and substitute the known conditions
F (1 + 0) = f (1) + F (0) + 2 * 1 * 0, that is, f (0) = 0,
Let y = - X and substitute it into the known condition
F (x-x) = f (x) + F (- x) + 2 * x * (- x)
f(-x)=2x^2-f(x) (1)
Let x = 1, y = 1 be substituted into known conditions
F (1 + 1) = f (1) + F (1) + 2 * 1 * 1 = 2 + 2 + 2 = 6, that is, f (2) = 6,
Let x = 2, y = 1 be substituted into known conditions
F (2 + 1) = f (2) + F (1) + 2 * 2 * 1 = 12, that is, f (3) = 12
So, let x = 3 be substituted into equation (1)
f(-3)=2*3^2-f(3)=18-12=6
My question is: why make x = 1, y = 0; y = - x; etc
What is the specific idea?

Starting from the two directions of the title and problem: if f (1) = 2, then assume that x + y = 1, who is 1 and who is 0, it doesn't matter. The key point is to get a new known condition f (0) = 0, and then deduce x + y = 0, we can get f (- x) = 2x ^ 2-F (x). At this time, we can see that f (- x) appears, which is the same form as the problem f (- 3)

If the function f (x) defined on R satisfies f (x + y) = f (x) + F (y) + 2XY (x, y belongs to R) f (1) = 2, then f (- 3) is urgent,

f(x+y)=f(x)+f(y)+2xy
f(2)=f(1)+f(1)+2=6
f(4)=f(2)+f(2)+8=20
f(1)=f(4)+f(-3)+2*4*(-3)=20+f(-3)-24=2
f(-3)=6

If f (x) defined on R satisfies f (x + y) = f (x) + F (y) + 2XY, f (1) = 2, then f (- 3)=
Why is f (0) = 0 when x = y = 0

When x = y = 0, f (0 + 0) = f (0) + F (0) + 0
That is, f (0) = 2F (0)
So f (0) = 0
f(-1+1)=f(0)=f(-1)+f(1)+2*(-1)*1=0
That is, f (- 1) = f (1) - 2 = 0
f(-1-1)=f(-2)=f(-1)+f(-1)+2*(-1)*(-1)=2f(-1)+2=2
f(-2-1)=f(-3)=f(-2)+f(-1)+2*(-2)*(-1)=f(-2)+f(-1)+4=6

It is proved that the limit of binary function does not exist

If the numerator and denominator are divided by XY at the same time, 1 / ((1 / x) + (1 / y)), 1 / Y - > ∞, the original formula becomes (x - > 0, Y - > ∞) limx - > ∞, so there is no limit

The method of proving the limit of binary function is not very clear

When f (x) x tends to infinity, the difference between the result and a certain number is 0,
F (x) / a = 1; f (x) - a = 0 and so on

How to prove the continuity and differentiability of functions
I want to know if we can prove that if we prove that LIM (x tends to 0) = f (0), we can say that it is continuous. If we prove that LIM (△ x tends to 0) = a real number, we can say that it is differentiable. Even if we can understand the process, we can't understand why it does so. I guess we can prove that it is continuous and differentiable according to these two ideas,

Continuity is as long as we prove that the left and right limits are equal and the function value at this point exists. The premise that a function can be derived at a certain point is that it is continuous at this point. After the continuity is known, we only need to prove that the left and right derivatives exist and are equal. The geometric meaning of the derivative is the slope of the tangent line of the curve represented by the function at this point

Why does the existence of the second derivative of a point mean that the first derivative is continuous in the field of the point, while the existence of the first derivative does not mean that the original function is continuous in the field of the point?
I see a lot of explanations: because the definition of the second derivative uses the first derivative, the first derivative is continuous at this point. So if the same first derivative exists at this point, why can't it show that the original function is continuous at this point? I understand all the examples, but I can't figure out the logic,
If the existence of the first derivative does not mean that the original function is continuous at this point, does the existence of the second derivative mean that the first derivative is continuous at this point?

Let f '' (x0) = Lim [f '(x) - f' (x0)] / (x-x0) exist, then Lim [f '(x) - f' (x0)] = 0. The above formula only shows that f '(x) is continuous in x = 0. Of course, it can show that f (x) is continuous in a neighborhood of x = 0

Continuous function is not necessarily differentiable, so why does continuous function have original function

First of all, continuous function must be integrable. This is a proved theorem. Here I just want to give a specific explanation. As for the proof of the theorem, please refer to the relevant textbooks. We know that the continuity, differentiability and integrability of functions are studied in calculus. But how strong are the three concepts of continuity, differentiability and integrability? We know that differentiability must be continuous, Note that these are derived in one direction (that is, not a necessary and sufficient condition), that is, there are some continuous functions but are not differentiable, and there are also some integrable functions but are not continuous, so we can say the strength of these three concepts: differentiable > continuous > integrable