张宇1000题高等数学 第四章 一元函数微分学的计算

目录

  • A A A
    • 11.设 y = ln ⁡ 1 − x 1 + x 2 y=\ln\sqrt{\cfrac{1-x}{1+x^2}} y=ln1+x21x ,求 y ′ ′ ∣ x = 0 y''\biggm\vert_{x=0} yx=0
  • B B B
    • 4.若 f ( x ) = x 5 e 6 x f(x)=x^5e^{6x} f(x)=x5e6x,则 f ( 101 ) ( 0 ) = f^{(101)}(0)= f(101)(0)=______。
    • 5.设 f ( x ) = x 1 − x 4 f(x)=\cfrac{x}{1-x^4} f(x)=1x4x,则 f ( 101 ) ( 0 ) = f^{(101)}(0)= f(101)(0)=______。
    • 16.设 f ( x ) = g ′ ( x ) , g ( x ) = { e x − 1 x , x ≠ 0 , 1 , x = 0 , f(x)=g'(x),g(x)=\begin{cases}\cfrac{e^x-1}{x},&x\ne0,\\1,&x=0,\end{cases} f(x)=g(x),g(x)=xex1,1,x=0,x=0, f ( n ) ( 0 ) f^{(n)}(0) f(n)(0)
  • C C C
    • 6.设 y = arcsin ⁡ x y=\arcsin x y=arcsinx
      • (1)证明其满足方程 ( 1 − x 2 ) y ( n + 2 ) − ( 2 n + 1 ) x y ( n + 1 ) − n 2 y ( n ) = 0 ( n ⩾ 0 ) (1-x^2)y^{(n+2)}-(2n+1)xy^{(n+1)}-n^2y^{(n)}=0(n\geqslant0) (1x2)y(n+2)(2n+1)xy(n+1)n2y(n)=0(n0)
      • (2)求 y ( n ) ∣ x = 0 y^{(n)}\biggm\vert_{x=0} y(n)x=0
    • 7.设 n n n为正整数, f ( x ) = ln ⁡ ( 1 + x 2 − x ) f(x)=\ln(\sqrt{1+x^2}-x) f(x)=ln(1+x2 x),求导数 f ( 2 n + 1 ) ( 0 ) f^{(2n+1)}(0) f(2n+1)(0)
  • 写在最后

A A A

11.设 y = ln ⁡ 1 − x 1 + x 2 y=\ln\sqrt{\cfrac{1-x}{1+x^2}} y=ln1+x21x ,求 y ′ ′ ∣ x = 0 y''\biggm\vert_{x=0} yx=0


y = ln ⁡ 1 − x 1 + x 2 = 1 2 [ ln ⁡ ( 1 − x ) − ln ⁡ ( 1 + x 2 ) ] , y ′ = 1 2 ( − 1 1 − x − 2 x 1 + x 2 ) = − 1 2 ( 1 1 − x + 2 x 1 + x 2 ) , y ′ ′ = − 1 2 [ 1 ( 1 − x ) 2 + 2 ⋅ 1 − x 2 ( 1 + x 2 ) 2 ] , y ′ ′ ∣ x = 0 = − 3 2 . \begin{aligned} &y=\ln\sqrt{\cfrac{1-x}{1+x^2}}=\cfrac{1}{2}[\ln(1-x)-\ln(1+x^2)],\\ &y'=\cfrac{1}{2}\left(\cfrac{-1}{1-x}-\cfrac{2x}{1+x^2}\right)=-\cfrac{1}{2}\left(\cfrac{1}{1-x}+\cfrac{2x}{1+x^2}\right),\\ &y''=-\cfrac{1}{2}\left[\cfrac{1}{(1-x)^2}+2\cdot\cfrac{1-x^2}{(1+x^2)^2}\right],\\ &y''\biggm\vert_{x=0}=-\cfrac{3}{2}. \end{aligned} y=ln1+x21x =21[ln(1x)ln(1+x2)],y=21(1x11+x22x)=21(1x1+1+x22x),y=21[(1x)21+2(1+x2)21x2],yx=0=23.
这道题主要利用了拆项求解

B B B

4.若 f ( x ) = x 5 e 6 x f(x)=x^5e^{6x} f(x)=x5e6x,则 f ( 101 ) ( 0 ) = f^{(101)}(0)= f(101)(0)=______。


f ( 101 ) ( 0 ) = ∑ k = 0 101 C 101 k ( x 5 ) ( k ) ( e 6 x ) ( 101 − k ) ∣ x = 0 = 5 ! 6 96 C 101 5 = 101 ! 96 ! 6 96 . \begin{aligned} f^{(101)}(0)&=\sum\limits_{k=0}^{101}\mathrm{C}^k_{101}(x^5)^{(k)}(e^{6x})^{(101-k)}\biggm\vert_{x=0}\\ &=5!6^{96}\mathrm{C}^5_{101}=\cfrac{101!}{96!}6^{96}. \end{aligned} f(101)(0)=k=0101C101k(x5)(k)(e6x)(101k)x=0=5!696C1015=96!101!696.
这道题主要利用了泰勒展开式求解

5.设 f ( x ) = x 1 − x 4 f(x)=\cfrac{x}{1-x^4} f(x)=1x4x,则 f ( 101 ) ( 0 ) = f^{(101)}(0)= f(101)(0)=______。

   f ( x ) = x 1 − 2 x 4 = x ∑ n = 0 ∞ ( 2 x 4 ) n = ∑ n = 0 ∞ 2 n x 4 n + 1 ( 2 x 4 < 1 ) f(x)=\cfrac{x}{1-2x^4}=x\sum\limits_{n=0}^\infty(2x^4)^n=\sum\limits_{n=0}^\infty2^nx^{4n+1}(2x^4<1) f(x)=12x4x=xn=0(2x4)n=n=02nx4n+1(2x4<1),又 f ( 101 ) ( 0 ) 101 ! = 2 25 \cfrac{f^{(101)}(0)}{101!}=2^{25} 101!f(101)(0)=225,则 f ( 101 ) ( 0 ) = 101 ! ⋅ 2 25 f^{(101)}(0)=101!\cdot2^{25} f(101)(0)=101!225。(这道题主要利用了泰勒展开式求解

16.设 f ( x ) = g ′ ( x ) , g ( x ) = { e x − 1 x , x ≠ 0 , 1 , x = 0 , f(x)=g'(x),g(x)=\begin{cases}\cfrac{e^x-1}{x},&x\ne0,\\1,&x=0,\end{cases} f(x)=g(x),g(x)=xex1,1,x=0,x=0, f ( n ) ( 0 ) f^{(n)}(0) f(n)(0)

  由泰勒展开式, e x − 1 x = 1 x [ ∑ n = 0 ∞ x n n ! − 1 ] = ∑ n = 1 ∞ x n − 1 n ! , x ≠ 0 \cfrac{e^x-1}{x}=\cfrac{1}{x}\left[\sum\limits_{n=0}^\infty\cfrac{x^n}{n!}-1\right]=\sum\limits_{n=1}^\infty\cfrac{x^{n-1}}{n!},x\ne0 xex1=x1[n=0n!xn1]=n=1n!xn1,x=0,且 ( ∑ n = 1 ∞ x n − 1 n ! ) ∣ x = 0 = 1 \left(\sum\limits_{n=1}^\infty\cfrac{x^{n-1}}{n!}\right)\biggm\vert_{x=0}=1 (n=1n!xn1)x=0=1,故
g ( x ) = ∑ n = 1 ∞ x n − 1 n ! = ∑ n = 0 ∞ x n ( n + 1 ) ! , f ( x ) = g ′ ( x ) = ∑ n = 1 ∞ n x n − 1 ( n + 1 ) ! = ∑ n = 0 ∞ ( n + 1 ) x n ( n + 2 ) ! . g(x)=\sum\limits_{n=1}^\infty\cfrac{x^{n-1}}{n!}=\sum\limits_{n=0}^\infty\cfrac{x^{n}}{(n+1)!},\\ f(x)=g'(x)=\sum\limits_{n=1}^\infty\cfrac{nx^{n-1}}{(n+1)!}=\sum\limits_{n=0}^\infty\cfrac{(n+1)x^{n}}{(n+2)!}. g(x)=n=1n!xn1=n=0(n+1)!xn,f(x)=g(x)=n=1(n+1)!nxn1=n=0(n+2)!(n+1)xn.
  根据展开式的唯一性,有 f ( n ) ( 0 ) n ! = n + 1 ( n + 2 ) ! \cfrac{f^{(n)}(0)}{n!}=\cfrac{n+1}{(n+2)!} n!f(n)(0)=(n+2)!n+1,故 f ( n ) ( 0 ) = 1 n + 2 ( n = 1 , 2 , ⋯   ) f^{(n)}(0)=\cfrac{1}{n+2}(n=1,2,\cdots) f(n)(0)=n+21(n=1,2,)。(这道题主要利用了泰勒展开式求解

C C C

6.设 y = arcsin ⁡ x y=\arcsin x y=arcsinx

(1)证明其满足方程 ( 1 − x 2 ) y ( n + 2 ) − ( 2 n + 1 ) x y ( n + 1 ) − n 2 y ( n ) = 0 ( n ⩾ 0 ) (1-x^2)y^{(n+2)}-(2n+1)xy^{(n+1)}-n^2y^{(n)}=0(n\geqslant0) (1x2)y(n+2)(2n+1)xy(n+1)n2y(n)=0(n0)

  由 y ′ = 1 1 − x 2 , y ′ ′ = x ( 1 − x 2 ) 3 2 y'=\cfrac{1}{\sqrt{1-x^2}},y''=\cfrac{x}{(1-x^2)^{\frac{3}{2}}} y=1x2 1,y=(1x2)23x,得 ( 1 − x 2 ) y ′ ′ − x y ′ = 0 (1-x^2)y''-xy'=0 (1x2)yxy=0。由莱布尼兹公式,有 ( 1 − x 2 ) y ( n + 2 ) − 2 n x y ( n + 1 ) − n ( n − 1 ) y ( n ) − x y ( n + 1 ) − n y ( n ) = 0 (1-x^2)y^{(n+2)}-2nxy^{(n+1)}-n(n-1)y^{(n)}-xy^{(n+1)}-ny^{(n)}=0 (1x2)y(n+2)2nxy(n+1)n(n1)y(n)xy(n+1)ny(n)=0,即 ( 1 − x 2 ) y ( n + 2 ) − ( 2 n + 1 ) x y ( n + 1 ) − n 2 y ( n ) = 0 ( n ⩾ 0 ) (1-x^2)y^{(n+2)}-(2n+1)xy^{(n+1)}-n^2y^{(n)}=0(n\geqslant0) (1x2)y(n+2)(2n+1)xy(n+1)n2y(n)=0(n0)。(这道题主要利用了特殊值求解

(2)求 y ( n ) ∣ x = 0 y^{(n)}\biggm\vert_{x=0} y(n)x=0

  在上式中令 x = 0 x=0 x=0,得 y ( n + 2 ) ( 0 ) = n 2 y ( n ) ( 0 ) ( n = 1 , 2 , ⋯   ) y^{(n+2)}(0)=n^2y^{(n)}(0)(n=1,2,\cdots) y(n+2)(0)=n2y(n)(0)(n=1,2,)。由于 y ( 0 ) ( 0 ) = y ( 0 ) = 0 y^{(0)}(0)=y(0)=0 y(0)(0)=y(0)=0,从而 y ( 2 k ) ( 0 ) = 0 y^{(2k)}(0)=0 y(2k)(0)=0。又因为 y ′ ( 0 ) = 1 y'(0)=1 y(0)=1,从而
y ( 2 k + 1 ) ( 0 ) = ( 2 k − 1 ) 2 y ( 2 k − 1 ) ( 0 ) = ⋯ = ( 2 k − 1 ) 2 ( 2 k − 3 ) 2 ⋯ 3 2 ⋅ 1 2 y ′ ( 0 ) = [ ( 2 k − 1 ) ! ! ] 2 ( k = 1 , 2 , ⋯   ) . \begin{aligned} y^{(2k+1)}(0)&=(2k-1)^2y^{(2k-1)}(0)=\cdots=(2k-1)^2(2k-3)^2\cdots3^2\cdot1^2y'(0)\\ &=[(2k-1)!!]^2(k=1,2,\cdots). \end{aligned} y(2k+1)(0)=(2k1)2y(2k1)(0)==(2k1)2(2k3)23212y(0)=[(2k1)!!]2(k=1,2,).
这道题主要利用了递推式求解

7.设 n n n为正整数, f ( x ) = ln ⁡ ( 1 + x 2 − x ) f(x)=\ln(\sqrt{1+x^2}-x) f(x)=ln(1+x2 x),求导数 f ( 2 n + 1 ) ( 0 ) f^{(2n+1)}(0) f(2n+1)(0)

  因为 f ′ ( x ) = 1 1 + x 2 − x ( x 1 + x 2 − 1 ) = − 1 1 + x 2 f'(x)=\cfrac{1}{\sqrt{1+x^2}-x}\left(\cfrac{x}{\sqrt{1+x^2}}-1\right)=-\cfrac{1}{\sqrt{1+x^2}} f(x)=1+x2 x1(1+x2 x1)=1+x2 1,所以 1 + x 2 f ′ ( x ) = − 1 , 1 + x 2 f ′ ′ ( x ) + x 1 + x 2 f ′ ( x ) = 0 \sqrt{1+x^2}f'(x)=-1,\sqrt{1+x^2}f''(x)+\cfrac{x}{\sqrt{1+x^2}}f'(x)=0 1+x2 f(x)=1,1+x2 f(x)+1+x2 xf(x)=0,即 ( 1 + x 2 ) f ′ ′ ( x ) + x f ′ ( x ) = 0 (1+x^2)f''(x)+xf'(x)=0 (1+x2)f(x)+xf(x)=0。方程两边对 x x x n − 1 n-1 n1次导,并利用莱布尼兹公式,得 ( 1 + x 2 ) f ( n + 1 ) ( x ) + 2 ( n − 1 ) x f ( n ) ( x ) + ( n − 1 ) ( n − 2 ) f ( n − 1 ) ( x ) + x f ( n ) ( x ) + ( n − 1 ) f ( n − 1 ) ( x ) = 0 (1+x^2)f^{(n+1)}(x)+2(n-1)xf^{(n)}(x)+(n-1)(n-2)f^{(n-1)}(x)+xf^{(n)}(x)+(n-1)f^{(n-1)}(x)=0 (1+x2)f(n+1)(x)+2(n1)xf(n)(x)+(n1)(n2)f(n1)(x)+xf(n)(x)+(n1)f(n1)(x)=0
  将 x = 0 x=0 x=0代入上式,得 f ( n + 1 ) ( 0 ) = − ( n − 1 ) 2 f ( n − 1 ) f^{(n+1)}(0)=-(n-1)^2f^{(n-1)} f(n+1)(0)=(n1)2f(n1)。把 n n n换为 2 n 2n 2n,并由此递推,得
f ( 2 n + 1 ) ( 0 ) = − ( 2 n − 1 ) 2 f ( 2 n − 1 ) ( 0 ) , f ( 2 n − 1 ) ( 0 ) = − ( 2 n − 3 ) 2 f ( 2 n − 3 ) ( 0 ) , ⋯ ⋯ f ′ ′ ′ ( 0 ) = − f ′ ( 0 ) = − 1. f^{(2n+1)}(0)=-(2n-1)^2f^{(2n-1)}(0),\\ f^{(2n-1)}(0)=-(2n-3)^2f^{(2n-3)}(0),\\ \cdots\cdots\\ f'''(0)=-f'(0)=-1. f(2n+1)(0)=(2n1)2f(2n1)(0),f(2n1)(0)=(2n3)2f(2n3)(0),f(0)=f(0)=1.
  由此,得 f ( 2 n + 1 ) ( 0 ) = ( − 1 ) n [ ( 2 n − 1 ) ! ! ] 2 f^{(2n+1)}(0)=(-1)^n[(2n-1)!!]^2 f(2n+1)(0)=(1)n[(2n1)!!]2。(这道题主要利用了构造方程求解

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