Find the derivative of the function: the process of y = 4x + 1!

The first, second, third and fourth floors are all right
On the fifth floor, the method is right, but the symbol is wrong:
D of DX represents infinitesimal, Δ In X Δ Is expressed as a finite small, which is strictly distinguished in calculus
Known: y = f (x) = 4x + 1
Try to calculate according to the definition: dy / DX
dy/dx = lim [f(x+ Δ x) - f(x)]/ Δ x
Δ x⟼0
= lim [4(x+ Δ x)+1 - (4x+1)]/ Δ x
Δ x⟼0
= lim [4 Δ x]/ Δ x
Δ ⟼0
= 4
The certificate is completed

Given that the definition field of the function f (x) = log3 [(MX2 + 8x + n) / (x2 + 1)] is R and the value field is [0,2], find the value of M and n Solution: let t = (MX2 + 8x + n) / (x2 + 1) then 1= Why is the domain of function f (x) = log3 [(MX2 + 8x + n) / (x2 + 1)] R So (1) there must be a real number solution, Then why is the discriminant greater than zero? Isn't there an intersection between the image and the x-axis?

It means that you can find some x and make the equation hold. Because you set t = (MX2 + 8x + n) / (x2 + 1), if the equation has no solution, you can't find t, but according to the known range of T is 1 to 9, which is contradictory, so there must be such a X
By the way, I don't know whether this solution is your own idea or taught by the teacher. The idea is very unclear and confusing. Here is a solution for your reference
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Known function f (x) = log3mx2 + 8x + n The definition field of x2 + 1 is r, the value field is [0,2], and find the value of M.N

Because f (x) = log3mx2 + 8x + n
The domain of x2 + 1 is r, ≓ x2 + 1 > 0, so MX2 + 8x + n > 0 is always true
Let y = MX2 + 8x + n
X2 + 1, since the value range of function f (x) is [0,2], then 1 ≤ y ≤ 9, and (y-m) • x2-8x + y-n = 0 holds
Since x ∈ R, ① if y-m ≠ 0, the discriminant △ = 64-4 (y-m) (y-n) ≥ 0, that is, Y2 - (M + n) y + mn-16 ≤ 0
Y = 1 and y = 9 are the two roots of equation Y2 - (M + n) y + mn-16 = 0,
When m + n = 10, mn-16 = 9, the solution is m = n = 5
② If y-m = 0, i.e. y = M = n = 5, the corresponding x = 0 meets the conditions
To sum up, M = n = 5

Given that the definition field of the function f (x) = log3 (MX ^ 2 + 8x + n) / (x ^ 2 + 1) is R and the value field is [0,2], find the values of M and n Baidu knows that there are some answers. I read some, but I can't understand one place: [solution I] F (x) = log3 ((MX ^ 2 + 8x + n) / (x ^ 2 + 1)) the value range is [0,2] So 0

The problem in solution 1; No contradiction! Because it is a quadratic function and its current definition domain is all real numbers, your question is not unreasonable. If the equation has a unique solution on (0, + ∞), your question is valuable; The only current solution is two overlapping; Problem in solution 2: judgment in solution 2

Known function f (x) = log3mx2 + 8x + n The definition field of x2 + 1 is r, the value field is [0,2], and find the value of M.N

Let f (x) = sin2xcos3x, find the n-th derivative of F (x) (n = 1,2,...)

The sum difference product formula and the n-order derivative formula of sinusoidal function are used. You can continue to ask me where you don't understand

Find the derivative of the function: the process of y = 4x + 1!