I'd like to ask how to prove that a function is differentiable in the domain of definition. It's better to have specific steps, and how to prove that a function is continuous in the domain of definition! It's in the domain, not at all!

Let's see the mathematical analysis published by East China Normal University. It's very clear. Generally, for proof, you need to use the definition to prove the derivative. The definition is that the ratio of the increment of function value △ y and the increment of independent variable △ x △ Y / △ X has a limit. This is where we say that f (x) is differentiable

The domain of the function f (x, y) = 1n (xsquare + ysquare-1) is

f(x,y)=ln[1/(x²+y²-1)]
The definition field is: X & # 178; + Y & # 178; - 1 > 0
That is: X & # 178; + Y & # 178; > 1
The image is the area outside the circle with the origin as the center and radius of 1

Sum function of power series (- 1) ^ (n-1) x ^ n / n

Because [(- 1) ^ (n-1) x ^ n / N] '= (- x) ^ (n-1)
So s' (x) = ∑ [(- 1) ^ (n-1) x ^ n / N] '= ∑ (- x) ^ (n-1) = 1 / (1 + x), - 1

Sum function of power series ∑ (1, + ∞) n (x-1) ^ n

∑n（x-1）^n=（x-1)∑n（x-1）^(n-1)
Let f (x) = ∑ n (x-1) ^ (n-1), integral term by term to get: ∫ [1, x] f (x) DX = ∫ [1, x] ∑ n (x-1) ^ (n-1) DX
=Σ (x) = ^ (1) / (x) = ^ - 1,
Therefore: ∑ n (x-1) ^ n = - (x-1) / x ^ 2

Sum function of power series ∑ (∞, n = 1) (- 1) ^ n (n + 1) x ^ n

Specific answers and explanations are as follows:
If you don't understand, please refer to the explanation below and pay attention to the concept of color correspondence

Power series ∑ (∞, n = 0) (n + 1) x ^ n / N!, | x|

Using the basic series expansion e ^ x = ∑ (∞, n = 0) x ^ n / N! To sum

Find the sum function of power function ∑ ((x + 1) ^ (2n-1)) / (n + 1). Note (0 - ∞)

The convergence domain is - 2 ≤ x < 0
Let t = (x + 1) & # 178; can we find it after transformation?

The monotone decreasing interval of the function y = log12 | x-3 | is______ ．

Let u = | x-3 |, then u is the decreasing function of X on (- ∞, 3), and u is the increasing function of X on (3, + ∞). And ∵ 0 ＜ 12 ＜ 1, y = log12u is the decreasing function of X on the interval (3, + ∞). So the answer is: (3, + ∞)

The monotone decreasing interval of function y = log (3) (1-x) + 1 / (1 + x) is______

Because 1-x > 0, that is X

What is the monotone decreasing interval of SiNx whose function f (x) = log is based on 2?

That is to find SiNx > 0 and the decreasing interval of SiNx
That is: 2K π + π / 2=

I'd like to ask how to prove that a function is differentiable in the domain of definition. It's better to have specific steps, and how to prove that a function is continuous in the domain of definition! It's in the domain, not at all!