What's the difference between functions of one variable and functions of many variables?

What's the difference between functions of one variable and functions of many variables?

Differentiable must be continuous, continuous is not necessarily differentiable!
Differentiable also must be continuous, continuous is not necessarily differentiable!
Functions of one variable are generally related to continuity and differentiability
Multivariate functions are generally related to differentiability and continuity

Why does a function have a tangent at a point but not necessarily differentiable at that point

If the tangent is perpendicular to the X axis, then the derivative is infinite, so it is not differentiable
For example, y = x ^ (1 / 3) is at x = 0

I have never understood what differential is and what kind of function is differentiable. Why are some multivariate functions differentiable but not differentiable

Differentiation, as the name implies, is infinite subdivision, that is, with the infinite subdivision of the independent variable, the dependent variable is also infinitely subdivided. The differentiability of a function is not the same as that of a certain point. Differentiability is generally only for a function. For a function, differentiability = differentiability = continuity + derivative exists everywhere. For a certain point, if it is not the endpoint, differentiability is basically equivalent to differentiability

Given y = x + UX + sin V, u = e ^ x, v = ln x, find dy / DX, find the solution

The answer is: dy / DX = 1 + e ^ x + X * (e ^ x) + cos (LNX) * (1 / x)
The derivation of X, UX and SINV is very simple. U * x uses the basic rule to get u * (the derivative of x) + X * (the derivative of U). SINV is a compound derivative. First, we take the derivative of SINV, and then multiply it by the derivative of V!
Welcome to ask!

General solution of differential equation (2x + e ^ y + 2) DX + e ^ y (x + 2E ^ Y-1) dy = 0
Not much

Should the title be in the form of mdx ndy = 0?
If the fruit belongs to the form of mdx ndy = 0, then
dM/dy=e^y,dN/dx=e^y
So let ∫ mdx = ∫ (2x + e ^ y + 2) DX = H (x, y) = f (x) + G (y)
That is ∫ (2x + e ^ y + 2) DX = (x ^ 2) + X (e ^ y) + 2x + G (y)
Then ^ Dy (y) = Dy (E) + (E)
So, G '(y) = 2 [e ^ (2Y)] - e ^ y
The solution is g (y) = ∫ {2 [e ^ (2Y)] - e ^ y} dy = [e ^ (2Y)] - (e ^ y) + C
To sum up, the general solution is
h(x,y)=(x^2)+x(e^y)+2x+[e^(2y)]-(e^y)+C
=(x-1)(e^y)+[e^(2y)]+(x^2)+2x+C

The range of y = 4sin x ^ 2-2 is?

y=4sin x^2-2
Range [- 6,2]
y=4(sin x)^2-2
Range [- 2,2]

Find the range of y = 4sinx + 4sin (2 π / 3 + x) + 2 roots 3

Y = 4sinx + 4sin (2 π / 3 + x) + 2 roots 3 = 4sinx + 4sin (π + X - π / 3) + 2 roots 3 = 4sinx + 4sin (π + x) cos (π / 3) - 4cos (π + x) sin (π / 3) + 2 roots 3 = 4sinx + 2Sin (π + x) - (2 roots 3) cos (π + x) + 2 roots 3 = 4sinx-2sinx + (2 roots 3) cosx + 2 roots 3 = 2sinx + (2 roots 3) cosx + 2 roots 3 = 4sin

The range of function y = 4cos ^ 2 + 4cos-2

Is it to find the range of y = 4cos & # x + 4cosx-2?
If so, the solution is as follows:
y=4cos²x+4cosx-2
=(2cosx+1)²-3
Because - 1 ≤ cosx ≤ 1
When cosx = - 1 / 2, the minimum value of the function is - 3; when cosx = 1, the maximum value of the function is 6
So the range of the function is [- 3,6]

7. Find the maximum value of the following functions and the set of independent variables X when the function reaches the maximum value: ① y = 1-1 / 2cosx ② y = 3sin (2x-2 / 3 π)

1) When cosx = - 1, y has a maximum of 3 / 2 {x | x = 2K π + π, K ∈ Z} when cosx = 1, y has a minimum of 1 / 2 {x | x = 2K π, K ∈ Z} 2) y has a maximum of 3 2x-2 π / 3 = 2K π + π / 2, x = k π + 7 π / 12 {x | x = k π + 7 π / 12 K ∈ Z} y has a minimum of - 3 2x-2 π / 3 = 2K - π / 2, x = k π + π / 12

If the function y = f (x) monotonically decreases in the interval [a, b], then the maximum value of F (x) is () and the minimum value is ()

If the function y = f (x) monotonically decreases in the interval [a, b], then the maximum value of F (x) is (f (a)), and the minimum value is (f (b))