Calculus

tags math calculus

Differential

f′(x)=lim⁡h→0f(x+h)+f(x)hf'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)+f(x)}{h} f′(x)=ddxf(x)f'(x)=\frac{d}{dx}f(x) ddxf(x)g(x)=f′(x)g(x)+f(x)g′(x)\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x) ddxf(x)g(x)=f′(x)g(x)+f(x)g′(x)g(x)2\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)+f(x)g'(x)}{g(x)^2} ddxf(g(x))=f′(g(x))×g′(x)\frac{d}{dx}f(g(x))=f'(g(x))\times g'(x)

example

ddxcos(x)=−sin(x)\frac{d}{dx}cos(x)=-sin(x) ddxtan(x)=sec(x)2\frac{d}{dx}tan(x)=sec(x)^2 ddxcot(x)=csc(x)2\frac{d}{dx}cot(x)=csc(x)^2 ddxsin−1(x)=11−x2\frac{d}{dx}sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}} ddxcos−1(x)=−11−x2\frac{d}{dx}cos^{-1}(x)=\frac{-1}{\sqrt{1-x^2}} ddxtan−1(x)=11+x2\frac{d}{dx}tan^{-1}(x)=\frac{1}{1+x^2}

intergrant

g′(x)=f(h(x))h′(x)−f(k(x))k′(x)g'(x)=f(h(x))h'(x)-f(k(x))k'(x) ∫f(x) dg(x)=f(x)g(x)−∫g(x) df(x)\int f(x) \ d g(x)=f(x)g(x)-\int g(x) \ d f(x)

example

∫ef(x) dx=ef(x)f′(x)+c\int e^{f(x)} \ dx=\frac{e^{f(x)}}{f'(x)}+c ∫ln⁡(x) dx=xln⁡(x)−x+c\int \ln(x) \ dx=x\ln(x)-x+c ∫1x dx=ln⁡(x)+c\int \frac{1}{x} \ dx=\ln(x)+c ∫sin(x) dx=−cos(x)+c\int sin(x) \ dx=-cos(x)+c ∫tan(x) dx=−ln⁡(cos(x))+c\int tan(x) \ dx=-\ln(cos(x))+c ∫sec(x)2 dx=tan(x)+c\int sec(x)^2 \ dx=tan(x)+c ∫2x1+x2 dx=ln⁡(1+x2)+c\int \frac{2x}{1+x^2} \ dx=\ln(1+x^2)+c

Taylor series 泰勒展開式

f(n)(a)=f(x)在x=a的n次導數f(x)=f(0)(a)0!(x−a)0+f(1)(a)1!(x−a)1+f(2)(a)2!(x−a)2⋯f(x)=∑n=0∞f(n)(a)n!(x−a)n% f(x)=a_0 x^0+a_1x^1+a_2x^2+a_3x^3 \cdots\\ % f'(x)=(1)a_1x^0+( 2 )a_2 x^1+(3 ) a_3 x^2\cdots\\ % f''(x)=( 2 )a_2 x^0+(6 ) a_3 x^1\cdots\\ f^{(n)}(a)=f(x)在x=a的n次導數\\ f(x)=\frac{f^{(0)}(a)}{0!}(x-a)^0+\frac{f^{(1)}(a)}{1!}(x-a)^1+\frac{f^{(2)}(a)}{2!}(x-a)^2 \cdots\\ f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n

find sin(x) Taylor series

f(x)=sin(x)→f0(x)=sin(x)f1(x)=cos(x)f2(x)=−sin(x)f3(x)=−cos(x)f4(x)=sin(x)=f0(x)if a=0f(x)=∑n=0∞f(n)(a)n!(x−a)n=∑n=0∞sin(0)(4n+0)!(x)4n+cos(0)(4n+1)!(x)4n+1+−sin(0)(4n+2)!(x)4n+2+−cos(0)(4n+3)!(x)4n+3=∑n=0∞cos(0)(4n+1)!(x)4n+1+−cos(0)(4n+3)!(x)4n+3=∑n=0∞1(4n+1)!(x)4n+1+−1(4n+3)!(x)4n+3sin(x)=∑n=0∞(−1)n(2n+1)!(x)2n+1 \begin{alignedat}{} f(x)=sin(x) \rightarrow &f^0(x)=sin(x)\\ & f^1(x)=cos(x)\\ & f^2(x)=-sin(x)\\ & f^3(x)=-cos(x)\\ & f^4(x)=sin(x)=f^0(x)\\ \end{alignedat}\\ \text{if }a=0\\ \begin{aligned}{} f(x)&=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\\ &=\sum_{n=0}^{\infty}\frac{sin(0)}{(4n+0)!}(x)^{4n}+\frac{cos(0)}{(4n+1)!}(x)^{4n+1}+\frac{-sin(0)}{(4n+2)!}(x)^{4n+2}+\frac{-cos(0)}{(4n+3)!}(x)^{4n+3}\\ &=\sum_{n=0}^{\infty}\frac{cos(0)}{(4n+1)!}(x)^{4n+1}+\frac{-cos(0)}{(4n+3)!}(x)^{4n+3}\\ &=\sum_{n=0}^{\infty}\frac{1}{(4n+1)!}(x)^{4n+1}+\frac{-1}{(4n+3)!}(x)^{4n+3}\\ sin(x)&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(x)^{2n+1}\\ \end{aligned}\\