It is known that ln y = (SiN x) * e ^ y

It is known that ln y = (SiN x) * e ^ y

Given, LN y = (SiN x) * e ^ y. find dy / DX
Solution 1: use the derivative formula of implicit function to solve the problem
Let f (x, y) = LNY - (SiNx) e ^ y = 0
Then dy / DX = - (&; F / &; x) / (&; F / &; y) = [(cosx) e ^ y] / [(1 / y) - (SiNx) e ^ y] = [y (cosx) e ^ y] / [1-y (SiNx) e ^ y]
Solution 2: direct derivation, pay attention to take y as an intermediate variable: [the trouble of this method is to solve y 'after derivation]
y'/y=(cosx)e^y+(sinx)(e^y)y'
So y '= y (cosx) e ^ y + y (SiNx) (e ^ y) y'
The result is [1-y (SiNx) e ^ y] y '= y (cosx) e ^ y
So y '= [y (cosx) e ^ y] / [1-y (SiNx) e ^ y]

When calculating the differential, Dy = ADX, a is the derivative. According to the definition, Dy = a Δ x, then DX = Δ X. but I think DX should be approximately equal to Δ X,

Why do you think it is approximately equal to Li? Let y = x (a function of first degree), DX can be understood as the differential of function y = x, DX = 1 * Δ x = Δ X. DX is called the differential of independent variable, and the expression of Δ X by DX is actually a simple form, and Δ x is the essence

Five teachers lead several students to travel (the travel fee is paid uniformly). They contact two travel agencies with the same price. After negotiation, the preferential terms of hotel a are: teachers pay in full, students pay 70% off; the preferential terms of travel agency B are: all teachers and students pay 80% off, Then which travel agency they choose will cost less

（1）：
Set the price to 1
5 teachers and some students, set the number of students as y
A travel agency:
7% discount for students, so: total price = 5 + 0.7y
B travel agency:
Students and teachers charge 20% off, so: total price = 0.8 (5 + y)
So when 5 + 0.7y = 0.8 (5 + y), so 5 + 0.7y = 4 + 0.8y, so y = 10
A: when the number of students is 10, the charge is the same
（2）：
Set the charge price as X
The charge of company a = 5x + 0.7xy
Company B's charge = 0.8x (5 + y)
If the number of students is less than 10, B travel agency is better
If the number of students is more than 10, a travel agency is better

Finding the value of x ^ x when x → 0 + is the limit Lim

y=x^x
lny=xlnx=lnx/(1/x)
Is ∞ / ∞ type, using the law of lobita
Molecular derivation = 1 / X
Denominator derivation = - 1 / X & sup2;
So = - x, the limit is 0
That is LIM (x → 0) LNY = 0
So LIM (x → 0) x ^ x = 1

5 (3A's Square b-ab) - (AB's square + 3A's Square b), where a = 1 / 2, B = 1 / 3 (grade 1)

5 (3a squared b-ab) - (AB squared + 3A squared b)
=Square B of 15a-5ab-ab-3a
=The square of 12a, the square of b-5ab-ab
If a = 1 / 2 and B = 1 / 3 are taken into account, the following results are obtained:
13-5/6-1/6
The solution is 12

I don't know how to say it
I saw some people say I don ` t konw what to say. I think how to say is right

I don ` t konw what to say
I don ` t konw how to say Chinglish
I don ` t konw what to say
The others are Chinglish, such as how to say, foreigners will laugh

The area of a classroom is about 50 square meters, and the area of several such classrooms is 1 hectare

1 ha = 10000 M2

If the power of 0 + (x + 1} of X is significant, then the value range of X is

It is meaningful that the power of 0 + (x + 1} of X is negative 2
∴﹛x≠0
x+1≠0
The value range of X is x ≠ 0 and X ≠ - 1

10 English words about hobbies
Not too much,

go fishing
watch TV
go shoping
go to the movie
go swimming
fly kites
read books
play sports
draw pictures
play computer games

{an} is an arithmetic sequence, the sum of the first n terms is Sn, and S4 = - 62, S6 = - 75, find | A1 | + | A2 | + +|Solution: let AK = 0
{an} is an arithmetic sequence, the sum of the first n terms is Sn, and S4 = - 62, S6 = - 75, find | A1 | + | A2 | + +|The value of A14 |
Solution: let AK = 0
Question: why is it set like this? Shouldn't AK be greater than a (K + 1)?

It can be seen from S4 = - 62, S6 = - 75 and {an} is arithmetic sequence
{an} is an increasing arithmetic sequence with A1 = - 20 d = 3
Because the title requires | A1 | + | A2 | + +|The value of A14 |
So we have to judge whether {an} is positive or negative first
To find the first positive number {an}
So let AK = 0