Let f (x) = Xe ^ x, then f (x) is polar___ Value for_______ .

Let f (x) = Xe ^ x, then f (x) is polar___ Value for_______ .

f(x)=xe^x
Then:
f'(x)=(x)'(e^x)＋(x)(e^x)'
f'(x)=(x＋1)e^x
The function f (x) decreases when it is (- 1, + ∞) and increases when it is (- 1, + ∞)
The function f (x) has a minimum, which is f (- 1) = - 1 / E

Let D: {(x, y) | x ^ 2 + y ^ 2 = 0}, f (x, y) be a continuous function on D, and f (x, y) = (1-x ^ 2-y ^ 2) ^ 1 / 2-8 / π f (U, V) dudv

Let f (x, y) be continuous, and f (x, y) = XY + ∫ d f (U, V) dudv, where D is y = 0, y = x 2, x = 1, then f (x, y) equals ()
What does ∫ f (U, V) dudv mean

The double integral ∫∫ d f (U, V) dudv and ∫∫ d f (x, y) DXDY are actually the same, only the letter has been changed. Obviously, in this formula, the double integral ∫∫ d f (U, V) dudv is calculated to obtain a constant. Let's set it as a, that is, f (x, y) = XY + a

Let f (x, y) be continuous and f (x, y) = XY + ∫∫ d f (U, V) dudv, where D is the region bounded by y = 0, y = X2, x = 1, then f (x, y) is equal to ()

Do double integrals at both ends at the same time. The integral at the right end of the equation is a number a, then a = the integral of XY on the region D + the area of a * region
Then solve the equation

Let d be a planar region bounded by y = 0, y = x ^ 2, x = 1, and f (x, y) = XY + ∫ ∫ (d) f (U, V) dudv, then f (x, y) =?

Let I = ∫ (d) f (U, V) dudv
The double integral of D is obtained from both sides of the original formula
I=+∫∫xydxdy+I∫∫dxdy
I can be solved by calculation

Let f (z) = u (x, y) + V (x, y) be analytic in domain D. It is proved that u (x, y) is also an analytic function in domain D
f(z)=u(x,y)+iv(x,y)

Let V (x, y) = 0
Or the partial derivatives of U (x, y) exist everywhere

Find the differential f (x) = x ^ X of this function
X is a positive real number

f(x) = x^x
lnf(x) = xlnx
(1/f(x)) f'(x) = 1+ lnx
f'(x) = (1+lnx)x^x

Finding the differential of the function y = f (x) determined by tany = x + y

Differential equation on both sides at the same time (equivalent to derivation) is: SecY square times y '= 1 + y‘
So y '= the square of Coty

The differential of function f (x) = xarctanx-1 / 2ln (1 + x ^ 3)

f(x)=xarctanx-1/2ln(1+x^3)
df(x)/dx=arctanx+x*1/(1+x^2)-1/2 *1/(1+x^3)*3x^2
=arctanx+x/(1+x^2)-3x^2/2(1+x^3)

When f (x) = f (A-X), please prove that the integral of function f (x) in [0, a] is equal to twice the integral of function f (x) in [0, a / 2]

1) integral [0, a] f (x) DX = integral [0, a / 2] f (x) DX + integral [A / 2, a] f (x) DX2) let x = a-y, then DX = - Dy, integral [A / 2, a] f (x) DX = integral [A / 2,0] f (a-y) (- dy) = integral [0, a / 2] f (a-y) dy = integral [A / 2, a] f (y) Dy3) integral [0, a