Y is equal to the x power of a under the root sign to find its derivative

Y is equal to the x power of a under the root sign to find its derivative

y=√a^x
=a^(x/2)
y'=lna*(1/2)*a^(x/2)

Y = x / root sign (1 + x ^ 2) for second derivative

How to prove monotonicity of check function with derivative
I think it's too troublesome to prove the monotonicity of function with domain of definition. Who can teach me to prove it with derivative, especially the check function (there are also Nike functions with different names)

When the derivative is greater than 0, the function increases monotonically, when the derivative is less than 0, the function decreases monotonically

Fish is the plural fish or fish

Fish for the same kind of complex number
Fish for different species

Is the plural of fish fish fish or fish?

Fish refers to a lot of fish
Fish is a lot of fish

Is fish plural?
Besides fish on the table, is fish plural?
If there are many fish in my home, can you say "there are many fish in my home"?

Fish plural generally means many kinds of fish, more than one species
A lot of fish, the meaning here is a large number, its singular and plural are in the same shape
There are many fish in my home

It is proved that an upper triangular matrix over the number field F must be similar to a lower triangular matrix

It can be verified that J ^ (- 1) = J, J left multiplication matrix A is equivalent to inverting a according to the horizontal symmetry axis, that is, the first row and the nth row, the second row and the Nth-1 row,... J right multiplication matrix A is equivalent to inverting a according to the vertical symmetry axis, that is, the first column and the nth column, the second column and the nth -

The matrix ab-ba = a over complex field is proved to have only zero eigenvalues

From ab-ba = a, there is ab = Ba + a = (B + e) a
Then there are ab & # 178; = (B + e) AB = (B + e) &# 178; a, AB & # 179; = (B + e) &# 179; a,., AB ^ k = (B + e) ^ k · a
In general, for any polynomial f (x), AF (b) = f (B + e) a can be obtained
Furthermore, we can get: A & # 178; f (b) = AF (B + e) a = f (B + 2e) a & # 178;, a & # 179; f (b) = f (B + 3e) a & # 179;,..., a ^ k · f (b) = f (B + Ke) a ^ K
If there is a polynomial f (x) and a positive integer k, f (b) = 0 and f (B + Ke) is invertible
Then from F (B + Ke) a ^ k = a ^ k · f (b) = 0, we can get a ^ k = 0, that is, a is nilpotent matrix
The nilpotent matrix has only zero eigenvalues (if λ is the eigenvalue of a, then λ ^ k is the eigenvalue of a ^ k, and a ^ k = 0 has λ = 0)
So we only need to find the polynomial f (x) and the positive integer K
According to Hamilton Cayley theorem, if f (x) is the characteristic polynomial of B, f (b) = 0
Let the eigenvalues of B, that is, the n roots (multiplicity) of F (x) be λ 1, λ 2,..., λ n
It is easy to see that there exists a positive integer k such that λ 1 + K, λ 2 + K,..., λ n + k are not roots of F (x)
So the eigenvalues f (λ 1 + k), f (λ 2 + k),..., f (λ n + k) of F (B + Ke) are not zero, that is, f (B + Ke) is reversible
In this way, f (x) and K satisfy the requirements
Note: finally, the conclusion is used here
If the eigenvalues of B are λ 1, λ 2,..., λ N and f (x) are arbitrary polynomials, then the eigenvalues of F (b) are f (λ 1), f (λ 2),..., f (λ n)
If B is similar to upper triangular matrix (such as Jordan canonical form), it can be proved only for upper triangular matrix, but it is obvious for upper triangular matrix

It is known that the complex numbers α and β satisfy | α | = 1, | β | = 1. It is proved by geometric method that | (α - β) / (1 - α β) | = 1

The title of the building is wrong. It's wrong to bring two values in

If the vector AC corresponds to the complex number 6 + 8i and the vector BD corresponds to the complex number - 4 + 6I in the parallelogram ABCD on the complex plane, then the complex number corresponding to the vector Da is____________ ?

Let vector AB correspond to complex number a + bi, and vector ad correspond to complex number x + Yi, then vector AC = vector AB + vector ad = (a + x) + (B + y) I = 6 + 8i, ｜ a + x = 6, B + y = 8, vector BD = vector ad, vector AB = (x-a) + (y-b) I, ｜ x-a = - 4, y-b = 6, and the solution is x = 1, y = 4, vector Da = - vector ad = - 1-4i