[Advanced Mathematics] Continuously Differentiable and Differentiable (Definition + Proof + Memory Method)

Continuously differentiable and differentiable

1.Definition

1.1 Definition of continuous

1. Definition 1

   Let y = f(x) at point $x_0$has a definition in a certain field, if

   $\underset{\Delta x\to 0}{\lim }\Delta y\_=\underset{\Delta x\to 0}{\lim }[f(x_0+\Delta x)-f(x_0)]=0$

   Then it is said that y = f(x) is at point $x_0$Continuous everywhere

2. Definition 2

   Let y = f(x) at point $x_0$has a definition in a certain field, if

   $\underset{\Delta x\to 0}{\lim }f(x)=f(x_0)$

   Then it is said that y = f(x) is at $x_0$Continuous everywhere

1.2 Definition of differentiable

Let y = f(x) at point $x_0$is defined in a certain field, if the limit

$\underset{\Delta x\to 0}{\lim }\frac{\Delta y}{\Delta x\_}=\underset{\Delta x\to 0}{\lim }\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x\_}$

exists, then f(x) is said to be at point $x_0$It is differentiable everywhere, denoted as $f^{\prime }(x_0)$ can also be written as$y^{\prime }{|}_{x=x_0},\frac{dy}{dx}{|}_{x=x_0},\frac{df(x)}{dx}{|}_{x=x_0}$

1.3 Definition of differentiable

Let y = f(x) at point $x_0$It is defined in a certain field $\Delta y\_=f(x_0+\Delta x)-f\left(x_0\right)$ can be expressed as

$\Delta y\_=A\Delta x+o(\Delta x),(\Delta x\_\to 0)$,

where A is independent of $\Delta x$ is a constant, then the function is said to be at$x$$x_0$Differentiable everywhere, and $A\Delta x$ is called the function$y=f\left(x\right)$ at point$x_0$Relative to the independent variable increment $The\; differential\; of\; \Delta x$ is denoted as dy, that is,
$dy=A\Delta x$

2. The relationship between the three

3. Proof of relationship

3.1 Differentiability and Differentiability

1. Differentiable must be conductive

   According to the above definition, assuming that the function is differentiable at the point y=f(x), divide both sides of the function by $\Delta x$ , we get

   $\frac{\Delta y}{\Delta x\_}=A+\frac{o(\Delta x)}{\Delta x\_}$

   当$\Delta x\_\to At\; 0$ o'clock, there is

   $A=\underset{\Delta x\to 0}{\lim }\frac{\Delta y}{\Delta x\_}=f^{\prime }(x_0)$

   Meet the definition of derivable

2. Derivable must be differentiable

   若$y=f(x_0)$ at point$x_0$can be derived, then

   $\underset{\Delta x\to 0}{\lim }\frac{\Delta y}{\Delta x\_}=f^{\prime }\left(x_0\right)$

   Exist, according to the relationship between limit and infinitesimal (Theorem 1, Chapter 1, Section 4, Advanced Mathematics), it can be written as

   $\frac{\Delta y}{\Delta x\_}=f^{\prime }\left(x_0\right)+a$

   where $a\to 0(when\Delta x\to 0)$ , multiply both sides by$\Delta x$ , there is

   $\Delta y\_=f^{\prime }(x_0)D+aDx$

   因$aDx=o\left(\Delta x\right)$ , and$f^{\prime }\left(x_0\right)$ does not depend on$\Delta x$ , consistent with the definition of differentiable

3.2 Differentiable and continuous

1. Differentiable (differentiable) must be continuous

   Tips: It can be seen from the above proof that the function $f(x_0)$ at$x_0$The necessary and sufficient condition for a point to be differentiable is that the function $f\left(x_0\right)$ at$x_0$Points are differentiable, so as long as it is proved that differentiability must be continuous, it can be deduced that differentiability must be continuous.

   The following proof can be derived that it must be continuous

   Let the function $y=f\left(x\right)$ is differentiable at x, that is

   $\underset{\Delta x\to 0}{\lim }\frac{\Delta y}{\Delta x\_}=f^{\prime }\left(x\right)$, according to the relationship between limit and infinitesimal (Theorem 1, Chapter 1, Section 4, Advanced Mathematics), it can be written as

   $\frac{\Delta y}{\Delta x\_}=f^{\prime }(x)+a$

   where $a\to 0(when\Delta x\to 0)$ , multiply both sides by$\Delta x$ , there is

   $\Delta y\_=f^{\prime }(x)\Delta x+aDx$

   It can be seen that when $\Delta x\_\to 0$时，$\Delta y\_\to 0$ , which conforms to the definition of function continuity 1, so the derivative must be continuous

2. Continuous is not necessarily differentiable

   Counterexample: $y=|x|continuous$ but not differentiable

   1. Proof 1

      The function image is as follows:

      

      The geometric meaning of the derivative is the slope of the tangent line, that is, $f^{\prime }(x_0)=tan\alpha$ , it is easy to know that the slope on the left side of the origin is -1 and on the right side is 1. The left and right derivatives are not equal, so they are not differentiable (the necessary and sufficient condition for derivation is that the left and right limits exist and are equal)

4.Memory method

Differentiability and differentiability are equivalent, so remember the relationship between the three. Just remember that continuous is not necessarily differentiable\differentiable .

5. Reference articles

1. Advanced Mathematics (7th Edition)
2. Wu Zhongxiang’s online course + basics of advanced mathematics

At the end of the article, I would like to thank Brother Huagong Yue for his help and guidance.