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math_ Derivative function derivation of limit & differential & derivative & derivative / logarithmic function (derivative definition limit method) / derivative formula derivation of exponential functi
2022-07-06 03:52:00 【xuchaoxin1375】
List of articles
differential & derivative & Wechat Business
Function in x = x 0 x=x_0 x=x0 Definition of derivative
- Define two points
A x 0 ( x 0 , f ( x 0 ) ) ; ( finger set x = x 0 It's about Of extremely limit ) B x = ( x , f ( x ) ) = ( x 0 + Δ x , f ( x 0 + Δ x ) ) { Δ x = x − x 0 Δ y = { f ( x ) − f ( x 0 ) f ( x 0 + Δ x ) − f ( x 0 ) x → x 0 * Δ x → 0 Yes when , also remember h = Δ x A_{x_0}(x_0,f(x_0));( Appoint x=x_0 Limit at ) \\ B_x=(x,f(x))=(x_0+\Delta x,f(x_0+\Delta x)) \\ \begin{cases} \Delta x=x-x_0 \\ \Delta y= \begin{cases} f(x)-f(x_0) \\ f(x_0+\Delta x)-f(x_0) \end{cases} \end{cases} \\ x\rightarrow x_0 \Longleftrightarrow \Delta x\rightarrow 0 \\ Sometimes , Also remember h=\Delta x Ax0(x0,f(x0));( finger set x=x0 It's about Of extremely limit )Bx=(x,f(x))=(x0+Δx,f(x0+Δx))⎩⎪⎨⎪⎧Δx=x−x0Δy={ f(x)−f(x0)f(x0+Δx)−f(x0)x→x0*Δx→0 Yes when , also remember h=Δx
lim Δ x → 0 Δ y Δ x = { lim Δ x → 0 f ( x 0 + Δ x ) − f ( x 0 ) Δ x lim x → x 0 f ( x ) − f ( x 0 ) x − x 0 lim h → 0 f ( x 0 + h ) − f ( x 0 ) h \lim_{\Delta x\rightarrow 0}{\frac{\Delta y}{\Delta x}} =\begin{cases} \lim\limits_{\Delta x\rightarrow 0}{\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}} \\ \lim\limits_{x\rightarrow x_0}{\frac{f(x)-f(x_0)}{x-x_0}} \\ \lim\limits_{h\rightarrow 0}{\frac{f(x_0+h)-f(x_0)}{h} } \end{cases} Δx→0limΔxΔy=⎩⎪⎪⎪⎨⎪⎪⎪⎧Δx→0limΔxf(x0+Δx)−f(x0)x→x0limx−x0f(x)−f(x0)h→0limhf(x0+h)−f(x0)
Usually , For the convenience of writing , The third form is often used for derivation :
f ′ ( x 0 ) = lim h → 0 f ( x 0 + h ) − f ( x 0 ) h f'(x_0)=\lim\limits_{ h \rightarrow 0}{\frac{f(x_0+h)-f(x_0)}{h} } f′(x0)=h→0limhf(x0+h)−f(x0)
Definition of derivative function
Similar to the definition of derivative , We define the derivative in x 0 x_0 x0 Replace with x
f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x = lim h → 0 f ( x + h ) − f ( x ) h f'(x)=\lim_{\Delta x\rightarrow0}\frac{f(x+\Delta x)-f(x)}{\Delta x} =\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} f′(x)=Δx→0limΔxf(x+Δx)−f(x)=h→0limhf(x+h)−f(x)
remember
g ( h ) = f ( x + h ) − f ( x ) h g(h)=\frac{f(x+h)-f(x)}{h} g(h)=hf(x+h)−f(x)be
f ′ ( x ) = lim Δ x → 0 g ( Δ x ) = lim h → 0 g ( h ) this in this Well Write , yes by 了 strong transfer , benefit use guide Count set The righteous seek guide Count ( guide Letter Count ) Of when Hou , By seek extremely limit Of Letter Count g ( h ) Of since change The amount h ( namely f ( x ) since change The amount x Of increase The amount Δ x ) And By seek guide Count Of f ( x ) Of since change The amount x No Same as f'(x)=\lim_{\Delta x \rightarrow 0}{g(\Delta x)} =\lim_{h \rightarrow 0}{g(h)} \\ This is what it says here , To emphasize , Use the definition of derivative to find the derivative ( Derivative function ) When , \\ The function whose limit is found g(h) The independent variable of h( namely f(x) The independent variables x The incremental \Delta x) And the derivative of f(x) The independent variable of x Different f′(x)=Δx→0limg(Δx)=h→0limg(h) this in this Well Write , yes by 了 strong transfer , benefit use guide Count set The righteous seek guide Count ( guide Letter Count ) Of when Hou , By seek extremely limit Of Letter Count g(h) Of since change The amount h( namely f(x) since change The amount x Of increase The amount Δx) And By seek guide Count Of f(x) Of since change The amount x No Same asobviously , f ( x ) stay x 0 It's about Of guide Count f ′ ( x 0 ) Just yes guide Letter Count f ′ ( x ) stay x = x 0 It's about Of Letter Count value f(x) stay x_0 Derivative at f'(x_0) Is the derivative function f'(x) stay x=x_0 The value of the function at f(x) stay x0 It's about Of guide Count f′(x0) Just yes guide Letter Count f′(x) stay x=x0 It's about Of Letter Count value
- f ′ ( x 0 ) = f ′ ( x ) ∣ x = x 0 = d x d y f'(x_0)=f'(x)|_{x=x_0}=\frac{dx}{dy} f′(x0)=f′(x)∣x=x0=dydx
Derivative derivation of logarithmic function ( Derivative definition limit method )
f ( x ) = l o g a x f ′ ( x ) = ( l o g a x ) ′ = lim h → 0 l o g a ( x + h ) − l o g a ( x ) h = lim h → 0 l o g a ( x + h x ) h = lim h → 0 1 h l o g a ( x + h x ) = lim h → 0 l o g a ( 1 + h x ) 1 h remember g ( h ) = l o g a ( 1 + h x ) 1 h ( l o g a x ) ′ = lim h → 0 g ( h ) ; g ( h ) Of since change The amount yes h ( g ( h ) take x see do often The amount ) The extremely limit yes 1 ∞ class type ; from The first Two heavy want extremely limit Of PUSH wide Male type have to To : A = lim h → 0 h x 1 h = 1 x the With Yes On u = ϕ ( h ) = ( 1 + h x ) 1 h ; u 0 = lim h → 0 u = e 1 x also from complex close Letter Count Of extremely limit shipment count Law be : lim h → 0 g ( h ) = lim u → u 0 l o g a u = l o g a u 0 = l o g a e 1 x root According to the in At the end of Male type have to To ( l o g a x ) ′ = l o g a e 1 x = ln e 1 x ln a = 1 x 1 ln a f(x)=log_a x \\ f'(x)=(log_a x)'=\lim_{h\rightarrow 0}\frac{log_a{(x+h)}-log_a(x)}{h} =\lim_{h\rightarrow 0}\frac{log_a(\frac{x+h}{x})}{h} \\=\lim_{h\rightarrow 0}\frac{1}{h}{log_a({x+h}{x})} \\=\lim_{h\rightarrow 0}{log_a{(1+\frac{h}{x})^{\frac{1}{h}}}} \\ remember g(h)={log_a{(1+\frac{h}{x})^{\frac{1}{h}}}} \\(log_a x)'=\lim_{h\rightarrow 0}g(h);g(h) The independent variable of is h(g(h) take x As a constant ) \\ The limit is 1^\infin type ; From the generalized formula of the second important limit :A=\lim_{h\rightarrow 0}\frac{h}{x}\frac{1}{h}=\frac{1}{x} \\ So for u=\phi(h)=(1+\frac{h}{x})^{\frac{1}{h}}; \\ u_0=\lim_{h\rightarrow 0}{u}=e^{\frac{1}{x}} \\ By the limit algorithm of composite function : \lim_{h\rightarrow 0}g(h)=\lim_{u\rightarrow u_0}log_a{u}=log_a u_0=log_a e^\frac{1}{x} \\ According to the bottom changing formula (log_a x)'=log_ae^{\frac{1}{x}}=\frac{\ln e^{\frac{1}{x}}}{\ln a}=\frac{1}{x}\frac{1}{\ln a} f(x)=logaxf′(x)=(logax)′=h→0limhloga(x+h)−loga(x)=h→0limhloga(xx+h)=h→0limh1loga(x+hx)=h→0limloga(1+xh)h1 remember g(h)=loga(1+xh)h1(logax)′=h→0limg(h);g(h) Of since change The amount yes h(g(h) take x see do often The amount ) The extremely limit yes 1∞ class type ; from The first Two heavy want extremely limit Of PUSH wide Male type have to To :A=h→0limxhh1=x1 the With Yes On u=ϕ(h)=(1+xh)h1;u0=h→0limu=ex1 also from complex close Letter Count Of extremely limit shipment count Law be :h→0limg(h)=u→u0limlogau=logau0=logaex1 root According to the in At the end of Male type have to To (logax)′=logaex1=lnalnex1=x1lna1
Derivative and differential
- Differential is another description of derivative
- derivative y ′ = d y d x y'=\frac{dy}{dx} y′=dxdy,( The differential of a function dy Divide by the argument x Differential of dx, So the derivative is also called Wechat Business )
Derivative of logarithmic function
( l o g a x ) ′ = 1 x ln a (log_ax)'=\frac{1}{x\ln a} (logax)′=xlna1
- The derivative of logarithmic function can be calculated by the second important limit
Derivation of inverse function
With a x Of guide Letter Count PUSH guide by example , benefit use back Letter Count seek guide Law be With a^x As an example , Using the derivation rule of inverse function With ax Of guide Letter Count PUSH guide by example , benefit use back Letter Count seek guide Law be
Direct functions
- x = x ( y ) = l o g a y x , y take value Fan around : y ∈ ( 0 , + ∞ ) x ∈ ( − ∞ , + ∞ ) ( since change The amount y Of ) Letter Count x Of guide Count : x ′ ( y ) = 1 y 1 ln a x=x(y)=log_ay \\x,y Value range : \\ y\in (0,+\infin) \\x \in (-\infin,+\infin) \\( The independent variables y Of ) function x The derivative of : \\x'(y)=\frac{1}{y}\frac{1}{\ln a} \\ x=x(y)=logayx,y take value Fan around :y∈(0,+∞)x∈(−∞,+∞)( since change The amount y Of ) Letter Count x Of guide Count :x′(y)=y1lna1
Inverse function
- y = y ( x ) = a x * Letter Count x ( y ) and Letter Count y ( x ) mutual by back Letter Count y=y(x)=a^x \\ \bigstar function x(y) And the function y(x) They're inverse functions to each other \\ y=y(x)=ax* Letter Count x(y) and Letter Count y(x) mutual by back Letter Count
Derivative of inverse function
- be : y ′ ( x ) = 1 x ′ ( y ) = 1 1 x ln a = x ln a namely , y ′ ( x ) = ( a x ) ′ = x ln a ∴ ( a x ) ′ = x ln a be : y'(x)=\frac{1}{x'(y)}=\frac{1}{\frac{1}{x\ln a}}=x\ln a \\ namely ,y'(x)=(a^x)'=x\ln a \\ \therefore (a^x)'=x\ln a be :y′(x)=x′(y)1=xlna11=xlna namely ,y′(x)=(ax)′=xlna∴(ax)′=xlna
Logarithmic derivation
With seek a x Of guide Letter Count by example , send use Yes Count seek guide Law ( " No benefit seek guide Law ) In order to a^x As an example , Use logarithmic derivation ( Bernoulli derivation ) With seek ax Of guide Letter Count by example , send use Yes Count seek guide Law ( " No benefit seek guide Law )
y = a x ln y = ln a x = x ln a two edge Same as when seek guide 1 y y ′ = ln a y ′ = y ln a = a x ln a namely , ( a x ) ′ = a x ln a y=a^x \\ \ln y=\ln a^x=x \ln a \\ Take derivatives on both sides at the same time \\ \frac{1}{y}y'=\ln a \\ y'=y\ln a=a^x \ln a \\ namely ,(a^x)'=a^x \ln a y=axlny=lnax=xlna two edge Same as when seek guide y1y′=lnay′=ylna=axlna namely ,(ax)′=axlna
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