Let y = y (x) be an implicit function determined by x ^ y = y ^ x, then dy / DX =?

Let y = y (x) be an implicit function determined by x ^ y = y ^ x, then dy / DX =?

Oh, the teacher talked about this problem today. Dy / DX = negative f (x) / F (y). The second method is to take the logarithm in on both sides and find the derivative of X, where y is y = y (x), so we need to use the derivative rule of composite function!

The process of finding the differential equation (1 + x) DX - (1-y) dy = 0

(1+x)dx-(1-y)dy=(dx-dy)+(xdx+ydy)=d(x-y)+d(x^2+y^2)=0
That is, D (X-Y) = - D (x ^ 2 + y ^ 2)
Integrate at both ends to get X-Y = - (x ^ 2 + y ^ 2) / 2 + C
So, (x ^ 2 + y ^ 2) / 2 + X-Y = C

The initial solution of the differential equation dy / DX + y = x satisfying the initial condition Y / x = 0 = 2

If the eigenvalue is - 1, the solution of Y '+ y = 0 is Y1 = CE ^ (- x)
Let the special solution be y * = ax + B, and substitute it into the original equation to obtain: a + ax + B = X
Comparison coefficient; a=1,a+b=0
The solution is a = 1, B = - 1
Therefore, the general solution is y = Y1 + y * = CE ^ (- x) + X-1
When x = 0, y = C-1 = 2, C = 3
So the solution is; y=3e^(-x)+x-1

A detailed explanation of the continuity and discontinuity of the function of a higher number

Firstly, f (x) is not defined at x = 0. To make the function continuous at this point, the limit of the function at this point must be equal to the defined function value
When x tends to 0, Cotx ~ 1 / TaNx ~ 1 / X (equivalent infinitesimal relation)
Then f (x) = (1-x) ^ (1 / x), take - x as t, then f (T) = (1 + T) ^ (- 1 / T)
Because the important limit (1 + T) ^ (1 / T) is = e when t tends to 0, f = 1 / E when t tends to 0
Then when x tends to 0, f (x) tends to 1 / E
If you still have questions, you can continue to ask~

What is the first type of discontinuity in the supervisor's high number?

A: the first type of discontinuity: there is a unilateral finite limit, including removable discontinuity and jumping discontinuity
The discontinuities that can be removed are: for example, defining functions
f(x)=
x. (x is not equal to 0);
2,(x=0),
It can be seen that the function f (x) is discontinuous at x = 0, but both the left and right limits exist. If x = 0, it is called a removable breakpoint
Jumping discontinuities are: for example, defining functions
f(x)=
x,(x<0)；
1+x,(x>=0),
It can be seen that the function is a piecewise function, which is discontinuous around x = 0, but the left limit is 0, the right limit is 1, and x = 0 is called the jump breakpoint
The second type of discontinuity is the point where there is no limit in the discontinuity field. For example, f (x) = 1 / x, x = 0 is the second type of discontinuity

F (x) = x / TaNx find the discontinuity of the function, and judge what kind of discontinuity it is

∵y=x/tanx
‡ x = k π, x = k π + π / 2 (k is an integer) is its discontinuity
∵ f (0 + 0) = f (0-0) = 1 (when k = 0)
Neither f (K π + 0) nor f (K π - 0) exists (when k ≠ 0)
f(kπ+π/2+0)=f(kπ+π/2-0)=0
‡ x = k π (is a non-zero integer) belongs to the second type of breakpoint,
X = 0 and x = k π + π / 2 (k is an integer) are removable breakpoints
Supplementary definition: when x = 0, y = 1. When x = k π + π / 2 (k is an integer), y = 0
The original function is continuous at points x = 0 and x = k π + π / 2 (k is an integer)
Firstly, the limit of denominator TaNx at - π / 2 and π / 2 does not exist; secondly, the limit of denominator TaNx (when x → 0) is equal to zero, which can not be said that the limit of function exists]
F (x) = x / TaNx has three discontinuities within (- π, π):
① X = 0, then the denominator is equal to zero;
② X = - π / 2, the denominator is not defined at this time;
③ X = π / 2, the denominator is not defined at this time
They are breakpoints that can be removed because:
①x→0,f(x)→1；
②x→-π/2,f(x)→0；
③x→π/2,f(x)→0.