How do I prove that a function is differentiable at some point? I'll prove continuity and differentiability. How do I prove differentiability

How do I prove that a function is differentiable at some point? I'll prove continuity and differentiability. How do I prove differentiability

If it is a univariate function, then differentiability and differentiability are equivalent, so we only need to prove differentiability. For multivariate functions, if differentiable, it must be differentiable, but if only the derivative function or directional derivative exists, it is not necessarily differentiable. If the directional derivative is continuous, it must be differentiable, as long as we prove that the directional derivative or partial derivative is continuous. Of course, another trick is to prove it by definition, Sometimes it has unexpected effects

Mathematical proof of function How to prove e>(1+1/x)^x That is, e is greater than 1 plus the x power of the sum of the reciprocal of X It's necessary to prove it, but I know it works and I don't know how to prove it... It's best to use a simple method

A condition is missing in your question: x > 0, otherwise it is not tenable. The following proof is given: 1. First "transform" the original question by reasoning: to prove the original question E > (1 + 1 / x) ^ x, that is, to prove ln (E) > x * ln (1 + 1 / x), that is, to prove 1 > x * ln (1 + 1 / x). Because x > 0, it is to prove: 1 / x > ln (1 + 1 / x). It is observed that there are

Prove to be a periodic function and find the analytical formula of the function When the even function f (x) with domain r satisfies f (x + 1) = - f (x), and X ∈ [- 1,1], f (x) = x ^ 2 Verification: 2 is a period of function f (x) Find the analytical formula of the function of F (x) on the interval [2k-1,2k + 1], K ∈ Z

It is proved that from F (x + 1 + 1) = - f (x + 1) = f (x) = f (x + 2), 2 is a cycle of F (x). Since x ∈ [- 1,0], f (x) = x ^ 2
Then, - x ∈ [0,1], f (- x) = x ^ 2
[2k-1,2k + 1] differs from [- 1,1] by 2K cycles. Therefore, in [2k-1,2k + 1], f (x) = x ^ 2

Proof of mathematical increasing function 1: It is proved that f (x) = X-1 / X is an increasing function in (- ∞, - 1) 2: It is proved that f (x) = x + 2 / X is an increasing function on the interval [root 2, + ∞]

1：
Let X1 and X2 belong to (- ∞, - 1) and X1 > x2
be
f（x1）-f(x2)
=x1-x2+1/x1-1/x2
=(x1-x2)*(1-1/x1x2)
Because X1 > X2, x1-x2 > 0
Because X1 and X2 belong to (- ∞, - 1), 1 / x1x20
So f (x1) - f (x2) > 0
Obtain certificate
2：
Let X1 and X2 belong to [root 2, + ∞] and X1 > x2
f(x1)-f(x2)
=x1-x2+2/x1-2/x2
=(x1-x2)*（1-2/x1x2）
Because X1 > X2, x1-x2 > 0
Because X1 and X2 belong to [root 2, + ∞], 2 / x1x2 = 0
So f (x1) - f (x2) > = 0
Obtain certificate

Is there a function that is continuous everywhere but non differentiable everywhere in the real function space? If it exists, the function is given; If not, please explain the reason

Piano function
F (x) = Σ [1 to ∞] A ^ n sin (b ^ n * x)
Where 0 < A

It is proved that the continuous function and monotone function on the measurable set E are measurable functions?

Let's first clarify the definition of measurable function. Let the function be f (x), then f is measurable, that is, if any real number T, e (F > t) (the subset on e that makes f > t) is measurable, then f is measurable function. This definition is adopted
① Continuous function, set as f. continuous function has one property: for any λ ∈ R, set {x | f (x) > λ \ X15} are all open sets. This is a theorem. You can see if it is in the book. If not, you can use the continuous function definition in mathematical analysis. Then for any real number T, e (F > t) is an open set. Of course, the open set is measurable, so F is measurable
② Monotone function, F. Let f monotonically increase and decrease. For any real number T, if t is in the value range of F, there must be a unique x0, so that f (x0) = t, so e (F > t) = interval (x0, + ∞) ∩ e, of course, is the angle of two measurable sets or measurable sets, so f is measurable. If t does not belong to the value range of F, take the number t0 closest to t but greater than t in the value range of F, f (x0) = T0, so
E (F > t) = interval (x0, + ∞) ∩ e is still a measurable set. If t is greater than any number in the value range, e (F > t) = ∅, of course, it is also measurable. In conclusion, the monotone function f is measurable