Find the logarithm of log base 2 10

Find the logarithm of log base 2 10

log2(10)=lg(10)/lg(2)=1/lg(2)=3.321928

Log is based on 1 and the logarithm of 1 is equal to 0, right By the way, why can't the base be 1

The power 0 of any number is equal to 1. Conversely, the logarithm of 1 of any base number is 0
Because any power of 1 is equal to 1, the logarithm of N whose base number is 1 is any number. Other numbers are meaningless except that the logarithm is 0 when n = 1

Log what is the logarithm of 2 with a base of 10

It is LG2, estimated to be about 0.30

Why is the reciprocal relationship between the derivative of the original function and that of the inverse function?

Y = y (x) primitive function derivative of primitive function: dy / DX
X = x (y) inverse function derivative of inverse function: DX / dy
It can be seen that DX / dy = 1 / (dy / DX)
That is, the derivative of the original function and the derivative of the inverse function are reciprocal
Example: original function y = Tan x
Inverse function x = arctan y
Derivative of primitive function dy / DX = sec? X
Derivative of inverse function DX / dy = 1 / (1 + y 2)
dx/dy = 1/(1+tan²x) = 1/sec²x = 1/(dy/dx)
That is, DX / dy and dy / DX are reciprocal

What is the relationship between the derivatives of two functions that are reciprocal functions? Let's explain y = 1 / X and y = LNX,

It doesn't matter. But ((f ^ - 1) (x)) '= 1 / (f (y))'. Note that the argument on the right is y

Let f (x) have second derivative, and f ′ (x) ≠ 0, x = g (y) and y = f (x) be inverse functions of each other. Let f ′ (x), f ″ (x) denote g ′ (y), G ″ (y)

It can be seen from the meaning of the title that:
X = g (y) and y = f (x) are inverse functions of each other,
Therefore, G ′ (y) = 1
f′(x)
g″(y)＝d
dyg′(y)＝d
DY1
f′(x)＝d
Dx1
f′(x)dx
dy＝−f″(x)
f′(x)21
f′(x)＝−f″(x)
f′(x)3

G (x) is the inverse function of monotone differentiable function f (x), f (1) = 2, f '(1) = - √ 3 / 3, then G' (2) =? Why don't you give me a reason?

-The reciprocal of 3 / 3
The derivative of F (x) is dy / DX
So the derivative of its inverse function is to swap X and y, which happens to be DX / Dy, and they are just derivatives
In this problem, x = 1 y = 2 dy / DX (x = 1, y = 2) = - √ 3 / 3
DX / Dy (x = 1, y = 2) is exactly the reciprocal of dy / DX

Let f (x) be differentiable, and f (x) derivative > 0, f (0) = 0, f (a) = B, G (x) be the inverse function of F (x), find ∫ f (x) DX (upper a, lower o) + ∫ g (x) DX (B, 0)

Changing the independent variable X of F (x) to y x is the function expression of G (x)
We know that f (x) is an increasing function and ∵ f (0) = 0
The integral over the interval (0. A) is positive
Similarly, the integral of G (x) on the interval (0. B) is positive
And the sum of area = a * B
Therefore, the result is a * B

Let y = f (x) be second derivative, and its first and second derivatives are not zero, and its inverse function is x = φ (y), then φ '' (y)=____

∵ if the inverse function of the function y = f (x) is x = φ (y), then under the condition that the inverse function is differentiable, we have φ '(y) = 1 / F' (x) ······ (*) assuming that (*) is differentiable, the right side of the equal sign is regarded as a fraction, and the two ends of the equation take the derivative of Y (y) = {- 1 / [f '(x)]} · [f' (x)] '(y)

Given that the derivative of F (2) = 4 f (x) is f '(x) and f' (2) = 5, the derivative of inverse function f (x) at x = 4 is obtained

Y = f (x), f (2) = 4, f '(2) = 5
If the inverse function is y = g (x), then G (4) = 2
F (g (x)) = x
The derivation of two sides to X is: F '(g (x)) g' (x) = 1
X = 4 into: F '(g (4)) g' (4) = 1
That is, f '(2) g' (4) = 1
5 g '(4) = 1
Therefore, G '(4) = 1 / 5