High school derivative calculation Function f (x) = e ^ X / (x-a) (where a < 0). If there is x ∈ (a, 0], so that f (x) ≤ 1 / 2, find the value range of A (1) Find the definition domain and monotone interval of F (x)

It's actually very simple
e^x/（x-a) ≤1/2
Because x-a > 0, move it directly to get
e^x ≤（x-a）/2
Taking both sides of the inequality sign as two functions, it is easy to get that both are monotonically increasing functions in the same coordinate system. For e ^ x, the maximum value of X ∈ (a, 0] is 1
The next is the key, because the title only requires "existence", so just make a value of (x-a) / 2 greater than or equal to e ^ X. so bring in the maximum value of (x-a) / 2, that is, when x = 0. Because if even its maximum value cannot be greater than e ^ x, it is "nonexistent"
So when x = 0
1≤-a/2
A ≤ - 2

Known derivation formula: (1 / x) '= - 1 / x ^ 2 There is a function f (x) = 9 / x, I convert it into 1 / (x / 9), and then bring it to the formula: F '(x) = [1 / (x / 9)]' = - 1 / (x ^ 2 / 81) = - 81 / x ^ 2, but it is wrong. The correct answer is - 9 / x ^ 2 What's wrong with my calculation? I can't find it all the time. Please point it out, In addition, is the derivative of a function with coefficients directly multiplied by the coefficient on the derivative? Why can't it be brought into the formula as a whole with x?

solution
f[x]=9/x
f'[x]=-9/x^2
This derivative can be like this
As 9x1 / X
So its derivative is
9x1/x'
This is how the derivative is obtained
I hope it will help you. I don't know how to ask questions

The more, the better. I want to practice calculation. Thank you

Go to the bookstore and buy a dictionary
The problems of derivative appear in piles
First, I'll give you some questions to quench your thirst:
1. Find the derivative of y = (cosx) ^ SiNx
2. Find the derivative of y = (1 + x) / SiNx
3. Find the second derivative of y = x [sin (LNX) + cos (INX)]
4. Find the second derivative of y = x (1 + x) ^ (1 / 2)
5. Find the derivative of y = [x + (x ^ 2 + 1) ^ (1 / 2)] ^ (1 / 2)

Find the derivative of SiN x / 2 times cos X / 2. I solve it as follows: SiN x / 2 times SiN x / 2 minus (COS X / 2 times cos X / 2) = - cos x, that is, using the multiplication principle of derivative But the correct answer is to convert the original formula into 1 / 2 SiN x and then find its derivative. What's the problem with my solution?

[SiN x / 2 times cos X / 2] '
=1/2cosx/2cosx/2＋sinx/2（-1/2sinx/2）
=1/2[cos ² x/2-sin ² x/2]
=1/2cosx
As it turned out, you made a mistake

Basic formula of high school derivative

Common derivative formula
1. Y = C (C is constant) y '= 0
2.y=x^n y'=nx^(n-1)
3.y=a^x y'=a^xlna
y=e^x y'=e^x
4.y=logax y'=﹙logae﹚/x
y=lnx y'=1/x
5.y=sinx y'=cosx
6.y=cosx y'=－sinx

High school derivative formula

8
(C)'=0
(x^n)'=nx^(n-1)
(sinx)'=cosx
(cosx)'=-sinx
(e^x)'=e^x
(lnx)'=1/x
(a^x)=a^xlna
(loga(x))'=1/(xlna)

High school derivative calculation Function f (x) = e ^ X / (x-a) (where a < 0). If there is x ∈ (a, 0], so that f (x) ≤ 1 / 2, find the value range of A (1) Find the definition domain and monotone interval of F (x)