张宇1000题高等数学 第三章 一元函数微分学的概念

目录

  • B B B
    • 6.设 f ( x ) f(x) f(x) ( − π 2 a , π 2 a ) ( a > 0 ) \left(-\cfrac{\pi}{2a},\cfrac{\pi}{2a}\right)(a>0) (2aπ,2aπ)(a>0)内有定义,且 f ′ ( 0 ) = a f'(0)=a f(0)=a,又对任意的 x , y , x + y ∈ ( − π 2 a , π 2 a ) x,y,x+y\in\left(-\cfrac{\pi}{2a},\cfrac{\pi}{2a}\right) x,y,x+y(2aπ,2aπ),有 f ( x + y ) = f ( x ) + f ( y ) 1 − f ( x ) f ( y ) f(x+y)=\cfrac{f(x)+f(y)}{1-f(x)f(y)} f(x+y)=1f(x)f(y)f(x)+f(y),求 f ( x ) f(x) f(x)
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B B B

6.设 f ( x ) f(x) f(x) ( − π 2 a , π 2 a ) ( a > 0 ) \left(-\cfrac{\pi}{2a},\cfrac{\pi}{2a}\right)(a>0) (2aπ,2aπ)(a>0)内有定义,且 f ′ ( 0 ) = a f'(0)=a f(0)=a,又对任意的 x , y , x + y ∈ ( − π 2 a , π 2 a ) x,y,x+y\in\left(-\cfrac{\pi}{2a},\cfrac{\pi}{2a}\right) x,y,x+y(2aπ,2aπ),有 f ( x + y ) = f ( x ) + f ( y ) 1 − f ( x ) f ( y ) f(x+y)=\cfrac{f(x)+f(y)}{1-f(x)f(y)} f(x+y)=1f(x)f(y)f(x)+f(y),求 f ( x ) f(x) f(x)

  由 f ( x + y ) = f ( x ) + f ( y ) 1 − f ( x ) f ( y ) f(x+y)=\cfrac{f(x)+f(y)}{1-f(x)f(y)} f(x+y)=1f(x)f(y)f(x)+f(y),令 y = 0 y=0 y=0,得 f ( 0 ) = 0 f(0)=0 f(0)=0,故
f ′ ( x ) = lim ⁡ Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x = lim ⁡ Δ x → 0 f ( Δ x ) + f ( x ) 1 − f ( x ) f ( Δ x ) − f ( x ) Δ x = lim ⁡ Δ x → 0 f ( Δ x ) [ 1 + f 2 ( x ) ] Δ x [ 1 − f ( x ) f ( Δ x ) ] = lim ⁡ Δ x → 0 f ( 0 + Δ x ) − f ( 0 ) Δ x ⋅ lim ⁡ Δ x → 0 1 + f 2 ( x ) 1 − f ( x ) f ( Δ x ) = f ′ ( 0 ) [ 1 + f 2 ( x ) ] . \begin{aligned} f'(x)&=\lim\limits_{\Delta x\to0}\cfrac{f(x+\Delta x)-f(x)}{\Delta x}\\ &=\lim\limits_{\Delta x\to0}\cfrac{\cfrac{f(\Delta x)+f(x)}{1-f(x)f(\Delta x)}-f(x)}{\Delta x}\\ &=\lim\limits_{\Delta x\to0}\cfrac{f(\Delta x)[1+f^2(x)]}{\Delta x[1-f(x)f(\Delta x)]}\\ &=\lim\limits_{\Delta x\to0}\cfrac{f(0+\Delta x)-f(0)}{\Delta x}\cdot\lim\limits_{\Delta x\to0}\cfrac{1+f^2(x)}{1-f(x)f(\Delta x)}\\ &=f'(0)[1+f^2(x)]. \end{aligned} f(x)=Δx0limΔxf(x+Δx)f(x)=Δx0limΔx1f(x)f(Δx)f(Δx)+f(x)f(x)=Δx0limΔx[1f(x)f(Δx)]f(Δx)[1+f2(x)]=Δx0limΔxf(0+Δx)f(0)Δx0lim1f(x)f(Δx)1+f2(x)=f(0)[1+f2(x)].
  所以 f ′ ( x ) 1 + f 2 ( x ) = a \cfrac{f'(x)}{1+f^2(x)}=a 1+f2(x)f(x)=a,即 d [ arctan ⁡ f ( x ) + C 1 ] d x = d ( a x + C 2 ) d x \cfrac{\mathrm{d}[\arctan f(x)+C_1]}{\mathrm{d}x}=\cfrac{\mathrm{d}(ax+C_2)}{\mathrm{d}x} dxd[arctanf(x)+C1]=dxd(ax+C2)。故有 arctan ⁡ f ( x ) + C 1 = a x + C 2 \arctan f(x)+C_1=ax+C_2 arctanf(x)+C1=ax+C2,即 arctan ⁡ f ( x ) = a x + C \arctan f(x)=ax+C arctanf(x)=ax+C
  由 f ( 0 ) = 0 f(0)=0 f(0)=0,得 C = 0 C=0 C=0,所以 f ( x ) = tan ⁡ a x ( a > 0 ) f(x)=\tan ax(a>0) f(x)=tanax(a>0)。(这道题主要利用了微分方程求解

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