Let y = f (x) be bounded and differentiable in (0, + ∞), then when limx → + ∞ f (x) = 0, there must be limx → + ∞ f '(x) = 0, The answer is an example of sin (x ^ 3) / X. does the derivative limit exist?

You can only prove that the derivative can be any value. It is impossible to get the conclusion that the derivative is 0

Given that limx - > π f (x) exists and f (x) = SiNx / (x - π) + 2limx - > π f (x), find limx - > π f (x) =?
It is known that limx - > π f (x) exists, and f (x) = SiNx / (x - π) + 2limx - > π f (x),
Find limx - > π f (x) =?

Simple, let the result limx - > π f (x) = a, a is the answer, a is a constant
So f (x) = SiNx / (x - π) + 2A. The limit of limx - > π is taken on both sides of this equation
Then limx - > π f (x) = limx - > π (SiNx / (x - π) + 2a)
Note that this formula is a = limx - > π SiNx / (x - π) + 2A
So we get a = (- 1) + 2A. Solving this simple equation, we get a = 1
So the answer is 1

It is proved by definition that the function f (x) = x2 + 2x - 1 is a decreasing function on (0, 1)

It is proved that: let X1 < X2, and x1, X2 ∈ (0, 1], then f (x1) - f (x2) = X12 + 2x1-1-2x2-1 = (X21 − X22) + 2 (1x1 − 1x2) = (x2-x1) [2x1x2 - (x1 + x2)] ∧ x1, X2 ∈ (0, 1], and X1 < X2, ∧ x2-x1 > 0, X1 + x2 < 2, 2x1x2 > 2 ∧ (x2-x1) [2x1x2 -...]

Finding y '- y = cosx, x = 0, y = 0 of differential equation

Let y = asinx + bcosx
y'=acosx-bsinx
y'-y=(acosx-bsinx)-(asinx+bcosx)
=(-a-b)sinx+(a-b)cosx=cosx
By comparing the coefficients of corresponding terms, we get - A-B = 0, A-B = 1
The solution is a = 1 / 2, B = - 1 / 2
So the special solution y = (1 / 2) * SiNx - (1 / 2) * cosx

A special solution of differential equation y '' + y = cosx y * =?

y=(xsinx)/2
Because the characteristic equation of Y '' + y = cosx is that R * r + 1 = 0 has a root of ± I, and this root conflicts with the following cosx = (e ^ (IX) + e ^ (- IX)) / 2 (see the book), so we need to set y = bxsinx, and then substitute it to find the undetermined coefficient: B = 1 / 2

(xcosy + cosx) y '= ysinx siny is this a total differential equation?

(xcosy+cosx)y'=ysinx-siny
xcosydy+cosxdy=ysinxdx-sinydx
xdsiny+sinydx=-ydcosx-cosxdy
dxsiny=-d(ycosx)
xsiny= -ycosx+C

A special solution to the differential equation y '+ ycotx = 5E ^ cosx satisfying the condition y (π / 2) = - 4

First, we solve the general solution of Y '+ ycotx = 0 = = > dy / y + cosxdx / SiNx = 0 = = > dy / y + D (SiNx) / SiNx = 0 = = > ln ι y ι ln ι SiNx = ln ι C ι (C is an integral constant) = = = > ysinx = C ι y' + ycotx = 0, and the general solution of Y '+ ycotx = 5E ^ (cosx) is y = C (x) /

Find the special solution of y = 1 when y '+ Y / x = cosx / X satisfies the condition x = π

∵y'+y/x=cosx/x==>xy'+y=cosx
==>xdy+ydx=cosxdx
==>d(xy)=d(sinx)
Ψ xy = SiNx + C (C is an integral constant)
∵ when the differential equation satisfies the condition x = π, y = 1
∴π*1=sinπ+C==>C=π
So the solution of the original equation is: xy = SiNx + π

How to solve the differential equation XYY '- y * y + 3x * x * x * cosx = 0?
Y 'is the derivative of Y over X
Denotes multiplication
yy=x*x(C-6sinx)

Let t = Y / x, y '= t + XT'
The original equation becomes: t × (T + T ') - t × T + 3xcosx = 0
That is TT '+ 3cosx = 0
d(t*t)/2dx=-3cosx
t*t=c-6sinx
That is, (Y / x) * * 2 = c-6sinx
Finally, y * y = x * x (c-6sinx)
OK?

What is the tangent equation passing through a point a (2, - 2) on the curve s: y = 3x-x ^ 3
Solution: let the tangent point be (a, 3a-a & # 179;)
y'=3-3x²
k=3-3a²
The tangent is y-3a + A & # 179; = (3-3a & # 178;) (x-a)
Over (2, - 2)
-2-3a+a³=(3-3a²)(2-a)
a³-3a²+4=0
A = - 1 or a = 2
So there are two tangents
A = - 1, tangent y = - 2
A = 2, tangent y = - 9x + 16
How to find the cubic equation of one variable here

a³-3a²+4=0
(a³-4a²+4a)+(a²-4a+4)=0
a(a²-4a+4)+(a²-4a+4)=0
(a+1)(a²-4a+4)=0
(a+1)(a-2)²=0
So a = - 1 or a = 2
Sometimes it is necessary to test the feeling to do this equation of higher degree
Here is an informal but effective method
First, find out all the coefficients of the equation (without sign)
1 3 4
Then the divisors of these coefficients are obtained
1 2 3 4
Then, the coefficients or the opposite numbers of the coefficients are substituted into the equation. Usually, we can find a way to make the equation hold
For example, 2 makes the equation true
So you can write a factor (A-2)
Then divide a & # 179; - 3A & # 178; + 4 by A-2
When doing this step, the expression to be divided should be completed. For example, this expression lacks a term of A
So it should be added as: A & # 179; - 3A & # 178; + 0A + 4, and then divide
a² -a -2
-------------------
a-2 |a³-3a²+0a+4
a³-2a²
--------------
-a²+0a
-a²+2a
-----------
-2a+4
-2a+4
----------
0
So a & # 179; - 3A & # 178; + 4 = (A-2) (A & # 178; - A-2) = (A-2) (A-2) (a + 1) = (A-2) &# 178; (a + 1)
In this way, the equation is divided into (A-2) &# 178; (a + 1) = 0
I don't know if you can understand ~ ~ ~ please ask me if you have any questions~~~~~

Let y = f (x) be bounded and differentiable in (0, + ∞), then when limx → + ∞ f (x) = 0, there must be limx → + ∞ f '(x) = 0, The answer is an example of sin (x ^ 3) / X. does the derivative limit exist?