How to Compute The Derivatives （如何求导数）（TBC）

$\text{A video by 3Blue1Brown in Bilibili}$A video by 3Blue1Brown in Bilibili


$\text{I don't have a systematic face to Derivatives before, but now I do.}$I don’t have a systematic face to Derivatives before, but now I do.

$\text{The Derivative of a function is a function which equals to the slope of that}$The Derivative of a function is a function which equals to the slope of that$\text{function's graph.}$function’s graph.

$\text{1. The Derivative of Power Functions}$1. The Derivative of Power Functions

$\text{Let's look at some simples.}$Let’s look at some simples.

$\text{We can treat }y=x^2\text{ to a square with side length }x.$We can treat y=x2 to a square with side length x.

$\text{Let the side length becomes to }(x+dx)\text{ so that the area increases }(2x·dx+dx^2).$Let the side length becomes to (x+dx) so that the area increases (2x⋅dx+dx2).

$\text{Because of the increment }dx\text{ should be much smaller than }x,\text{ so it's safe to ignore the}$Because of the increment dx should be much smaller than x, so it’s safe to ignore the$\text{terms those have }dx^2\text{ or high power of }dx.$terms those have dx2 or high power of dx.
$\text{Therefore, }\Delta S=2x·dx.\text{ When }dx\rightarrow0,\Delta S\rightarrow 2x.$Therefore, ΔS=2x⋅dx. When dx→0,ΔS→2x.

$\text{And the same, consider the function }y=x^3.$And the same, consider the function y=x3.
$\text{We can treat it to a cube with a side length }x.$We can treat it to a cube with a side length x.

$\text{Let the side length becomes to }(x+dx)\text{ so that the volume increases }(3x^2·dx+3x·dx^2+dx^3).$Let the side length becomes to (x+dx) so that the volume increases (3x2⋅dx+3x⋅dx2+dx3).
$\text{As the same, we also let it be }3x^2.$As the same, we also let it be 3x2.


$\text{Actually, for a power function }f(x)=x^n,\text{ it goes}$Actually, for a power function f(x)=xn, it goes
$\begin{aligned}f(x)&=x^n\\\\ f(x+dx)&=(x+dx)^n\\ &=x^n+nx^{n-1}dx+...\\\\ f'=\Delta f&=f(x+dx)-f(x)\\ &=nx^{n-1}\end{aligned}$f(x)f(x+dx)f′=Δf​=xn=(x+dx)n=xn+nxn−1dx+...=f(x+dx)−f(x)=nxn−1​
$\text{Therefore, its dericatives }f'=nx^{n-1}.$Therefore, its dericatives f′=nxn−1.


$\text{To be continued.}$To be continued.