The derivative of negative X of E is equal to? Why, don't you understand?

The derivative of negative X of E is equal to? Why, don't you understand?

This is the derivative of a compound function. Let u = - X
The answer is minus X of - E

What is the derivative of e ^ 2 / x Such as title

（e^2/x）'=-e^2/(x^2)
(e^（2/x）)'=e^（2/x）*(-1/(x^2))
Note that the minus sign is not long enough=

Let f (x), G (x) have second-order continuous derivatives, and the curve integral ∮ (lower C) [y ^ 2F (x) + 2ye ^ x + 2yg (x)] DX + 2 [YG (x) + F (x)] dy = 0 Where C is any simple closed curve in the plane (1) Find f (x), G (x) so that f (0) = g (0) = 0 (2) Calculate the integral from (0,0) to (1,1) along any curve

∮ (lower C) [y ^ 2F (x) + 2ye ^ x + 2yg (x)] DX + 2 [YG (x) + F (x)] dy = 0
obvious
Can get
f(x)=g(x)'     f(x)'-e^x-g(x)=0
Just solve the differential equation

Under the high number, if for any simple closed curve L on the plane, there is always ∮ YF (x) DX + [f (x) - x Λ 2] dy = 0, f (0) = 2, find f (x) If for any simple closed curve L on the plane, there is always ∮ YF (x) DX + [f (x) - x Λ 2] dy = 0, where f (x) has a continuous first derivative in (- ∞, ∞), and f (0) = 2, find f (x)

∮yf(x)dx+[f(x)-x∧2]dy=0
[yf(x)]'y=f(x)
[f(x)-x∧2]'x=f'(x)-2x
f'(x)-2x=f(x)
f'(x)=f(x)+2x
The general solution formula of first-order differential equation:
f(x)=Ce^x-2x-2
F (0) = 2 substituted into: C = 4
f(x)=4e^x-2x-2

It is proved that the curve integral is independent of the path: ∫ (x + y) DX + (X-Y) dy {upper limit of integral (2,3), lower line (1,1)} is in the whole xoy Prove that the curve integral is independent of the path: ∫ (x + y) DX + (X-Y) dy {the upper limit of the integral (2,3), and the lower line (1,1)} is independent of the path in the whole xoy plane. Calculate the score

∫ P dx+Q dy
To prove that this integral is independent of the path, just prove it ə Q/ ə x= ə P/ ə y
Let P = x + y, q = X-Y, then
ə Q/ ə x=1= ə P/ ə y
The curve integral is independent of the path (in the whole xoy plane)
‡ original integral = ∫ (x0, x1) P (x, Y0) DX + ∫ (Y0, Y1) Q (x1, y) dy
Or = ∫ (x0, x1) P (x, Y1) DX + ∫ (Y0, Y1) Q (x0, y) dy
For this question, there are
Original integral = ∫ (1,2) (x + 1) DX + ∫ (1,3) (2-y) dy
=[x ²/ 2+x]|(1,2)+[2y-y ²/ 2]|(1,3)
=5/2+0
=5/2
I hope my answer will help you
Don't forget to adopt it in time!

∮ τ (y-z)dx+(z-x)dy+(x-y)dz, τ It is an ellipse x ^ 2 + y ^ 2 = a ^ 2, X / A + Z / b = 1, if you go from the positive direction of the X axis ∮ τ (y-z)dx+(z-x)dy+(x-y)dz, τ It is an ellipse x ^ 2 + y ^ 2 = a ^ 2, X / A + Z / b = 1, (a > 0, b > 0). If you look at it from the positive direction of the X axis, the ellipse is counterclockwise

Use Stokes formula
P=y-z;
Q=z-x;
R=x-y;
Original formula = double integral (- 1-1) dydz + (- 1-1) dzdx + (- 1-1) DXDY
=-2 double integral (1dydz + 1dzdx + 1dxdy)
=-2 * (0 + ab π + A * a π) = - 2A π (a + b) I don't know if there is wood or miscalculation... If you calculate again, this is the method