[Advanced Mathematics Part of Postgraduate Entrance Examination] Derivative Derivation

There are principles for setting the denominator, and only simple factors are decentralized

n order derivation

( u v ) ( n ) = ∑ k = 0 n C n k u k v n − k = C n 0 u ( 0 ) v ( n ) + C n 1 u ( 1 ) v ( n − 1 ) + + C n n u ( n ) v ( 0 ) (uv)^{(n)}=\sum_{k=0}^nC_n^ku^{k}v^{n-k}\\ =C_n^0u^{(0)}v^{(n)}+C_n^1u^{(1)}v^{(n-1)}++C_n^nu^{(n)}v^{(0)} (uv)(n)=k=0nCnkukvnk=Cn0u(0)v(n)+Cn1u(1)v(n1)++Cnnu(n)v(0)

Guidance formula

Absolute value derivation blind ( xa ) ′ = axa − 1 ( a is a constant) ( ax ) ′ = axlna ( ex ) ′ = ex ( logax ) ′ = 1 xlna ( a > 0 , a ≠ 1 ) ( lnx ) ′ = 1 x ( ln ∣ x ∣ ) ′ = 1 x ( sinx ) ′ = cosx ( cosx ) ′ = − sinx ( arcsinx ) ′ = 1 1 − x 2 ( arccosx ) ′ = − 1 1 − x 2 ( tanx ) ′ = sec 2 x ( cotx ) ′ = − csc 2 x ( arctanx ) ′ = 1 1 + x 2 ( arccotx ) ′ = − 1 1 + x 2 ( secx ) ′ = secxtanx ( cscx ) ′ = − cscxtanx ( ln ∣ secx + tanx ∣ ) ′ = secx ( ln ∣ cscx − cotx ∣ ′ = cscx ( ln ( 1 + x 2 + 1 ) ) ′ = 1 x 2 + 1 ( ln ( 1 + x 2 − 1 ) ) ′ = 1 x 2 − 1 \color{red}{\text{Absolute value guide is blind}} \begin{array}{l} (x^a)'=ax^{a-1} (\text{a is a constant }) \quad \quad (a^x)'=a^xlna \quad \quad (e^x)'=e^x \\ \\ (log_ax)'=\frac{1}{xlna} (a> 0,a\neq1) \quad \quad (lnx)'=\frac{1}{x} \quad \quad (ln|x|)'=\frac{1}{x} \\ \\ (sinx)'=cosx \quad (cosx)'=-sinx \\ \\ (arcsinx)'=\frac{1}{\sqrt{1-x^2}} \quad \quad (arccosx)'=-\frac{1}{\sqrt{1-x^2}} \\ \\ (tanx)'=sec^2{x} \quad \quad (cotx)'=-csc^2{x} \\ \\ (arctanx)'=\frac{1}{1+x^2} \quad \quad (arccotx)'=-\frac{1}{1+x^2} \\ \\ (secx)'=secxtanx \quad \quad (cscx)'=-cscxtanx \\ \\ (ln|secx+tanx|)'=secx \quad \quad (ln|cscx-cotx|'=cscx \\ \\ (ln(1+\sqrt{x^2+1}))'=\frac{1}{\sqrt{x^2+1}} \quad \quad (ln(1+\sqrt{x^2-1}))'=\frac{1}{\sqrt{x^2-1}} \\ \end{array}\\ \\ (arctanx)'=\frac{1}{1+x^2} \quad \quad (arccotx)'=-\frac{1}{1+x^2} \\ \\ (secx)'=secxtanx \quad \quad (cscx)'=-cscxtanx \\ \\ (ln|secx+tanx|)'=secx \quad \quad (ln|cscx-cotx|'=cscx \\ \\ (ln(1+\sqrt{x^2+1}))'=\frac{1}{\sqrt{x^2+1}} \quad \quad (ln(1+\sqrt{x^2-1}))'=\frac{1}{\sqrt{x^2-1}} \\ \end{array}\\ \\ (arctanx)'=\frac{1}{1+x^2} \quad \quad (arccotx)'=-\frac{1}{1+x^2} \\ \\ (secx)'=secxtanx \quad \quad (cscx)'=-cscxtanx \\ \\ (ln|secx+tanx|)'=secx \quad \quad (ln|cscx-cotx|'=cscx \\ \\ (ln(1+\sqrt{x^2+1}))'=\frac{1}{\sqrt{x^2+1}} \quad \quad (ln(1+\sqrt{x^2-1}))'=\frac{1}{\sqrt{x^2-1}} \\ \end{array}Absolute Value Seeking Ignores(xa)=axa 1 (ais a constant)(ax)=ax lno(ex)=ex(logax)=x l no1(a>0,a=1)( l n x )=x1(lnx)=x1( s in x )=cos x( cos x )=- s in x(arcsinx)=1x2 1(arccosx)=1x2 1( t an x ​​)=sec2x _( co t x )=csc2x _(arctanx)=1+x21(arccotx)=1+x21( sec x )=secxtanx(cscx)=cscxtanx(lnsecx+tanx)=sec x(lncscxcotx=cscx(ln(1+x2+1 ))=x2+1 1(ln(1+x21 ))=x21 1

Derivative of n order derivative

( a x ) ( n ) = a x ( l n a ) n ( e x ) ( n ) = e x ( s i n k x ) ( n ) = k n s i n ( k x + π 2 n ) ( c o s k x ) ( n ) = k n c o s ( k x + π 2 n ) ( l n x ) ( n ) = ( − 1 ) ( n − 1 ) ( n − 1 ) ! x n ( x > 0 ) ( l n ( 1 + x ) ) ( n ) = ( − 1 ) n − 1 ( n − 1 ) ! ( 1 + x ) n ( x > − 1 ) [ ( x + x 0 ) m ] ( n ) = m ( m − 1 ) ( m − 2 ) … ( m − n + 1 ) ( x + x 0 ) m − n ( 1 x + a ) ( n ) = ( − 1 ) n n ! ( x + a ) n + 1 \begin{array}{l} \\ (a^x)^{(n)} = a^x(lna)^n \quad\quad (e^x)^{(n)}=e^x\\ \\ (sinkx)^{(n)} = k^nsin(kx+\frac{\pi}{2}n)\\ \\ (coskx)^{(n)} = k^ncos(kx+\frac{\pi}{2}n)\\ \\ (lnx)^{(n)} = (-1)^{(n-1)}\frac{(n-1)!}{x^n}\quad (x>0)\\ \\ (ln(1+x))^{(n)} = (-1)^{n-1}\frac{(n-1)!}{(1+x)^n}\quad(x>-1)\\ \\ [(x+x_0)^m]^{(n)}=m(m-1)(m-2)\dots(m-n+1)(x+x_0)^{m-n}\\ \\ (\frac{1}{x+a})^{(n)}=\frac{(-1)^nn!}{(x+a)^{n+1}}\\ \end{array} (ax)(n)=ax(lna)n(ex)(n)=ex(sinkx)(n)=kn sin(kx+2pn)(coskx)(n)=kncos(kx+2pn)( l n x )(n)=(1)(n1)xn(n1)!(x>0)(ln(1+x))(n)=(1)n1(1+x)n(n1)!(x>1)[(x+x0)m](n)=m(m1)(m2)(mn+1)(x+x0)mn(x+a1)(n)=(x+a)n+1(1)nn!

example

[ ( x − x 0 ) n ] ( n ) = n ! [(x-\color{red}{x_0}\color{black})^n]^{(n)}=n! [(xx0)n](n)=n!

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Origin blog.csdn.net/Little_Matches/article/details/122618578