【考研数学高数部分】导数求导

设置分母有原则,简单因式才下放

n阶求导

( 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)

求导公式

绝对值求导视而不见 ( x a ) ′ = a x a − 1 ( a为常数 ) ( a x ) ′ = a x l n a ( e x ) ′ = e x ( l o g a x ) ′ = 1 x l n a ( a > 0 , a ≠ 1 ) ( l n x ) ′ = 1 x ( l n ∣ x ∣ ) ′ = 1 x ( s i n x ) ′ = c o s x ( c o s x ) ′ = − s i n x ( a r c s i n x ) ′ = 1 1 − x 2 ( a r c c o s x ) ′ = − 1 1 − x 2 ( t a n x ) ′ = s e c 2 x ( c o t x ) ′ = − c s c 2 x ( a r c t a n x ) ′ = 1 1 + x 2 ( a r c c o t x ) ′ = − 1 1 + x 2 ( s e c x ) ′ = s e c x t a n x ( c s c x ) ′ = − c s c x t a n x ( l n ∣ s e c x + t a n x ∣ ) ′ = s e c x ( l n ∣ c s c x − c o t x ∣ ′ = c s c x ( l n ( 1 + x 2 + 1 ) ) ′ = 1 x 2 + 1 ( l n ( 1 + x 2 − 1 ) ) ′ = 1 x 2 − 1 \color{red}{\text{绝对值求导视而不见}} \begin{array}{l} (x^a)'=ax^{a-1} (\text{a为常数}) \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} 绝对值求导视而不见(xa)=axa1(a为常数)(ax)=axlna(ex)=ex(logax)=xlna1(a>0,a=1)(lnx)=x1(lnx)=x1(sinx)=cosx(cosx)=sinx(arcsinx)=1x2 1(arccosx)=1x2 1(tanx)=sec2x(cotx)=csc2x(arctanx)=1+x21(arccotx)=1+x21(secx)=secxtanx(cscx)=cscxtanx(lnsecx+tanx)=secx(lncscxcotx=cscx(ln(1+x2+1 ))=x2+1 1(ln(1+x21 ))=x21 1

n阶导数求导

( 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)=knsin(kx+2πn)(coskx)(n)=kncos(kx+2πn)(lnx)(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!

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

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转载自blog.csdn.net/Little_Matches/article/details/122618578