极限法求下列函数导数(用导数定义) y=sin(-X+1) Y=In(x^2+1)
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极限法求下列函数导数(用导数定义) y=sin(-X+1) Y=In(x^2+1)
极限法求下列函数导数(用导数定义) y=sin(-X+1) Y=In(x^2+1)
极限法求下列函数导数(用导数定义) y=sin(-X+1) Y=In(x^2+1)
我来!第一题:
第二题:
(Δx→0) lim {sin[1-(x+Δx)] - sin(1-x)} / (x+Δx-x)
= (Δx→0) lim 2 cos(1-x - Δx /2) sin(- Δx /2) / Δx
= - cos(1-x) * (Δx→0) lim sin(Δx /2) / (Δx/2)
= - cos(1-x)
(Δx→0) lim { ln[(x+Δx)...
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(Δx→0) lim {sin[1-(x+Δx)] - sin(1-x)} / (x+Δx-x)
= (Δx→0) lim 2 cos(1-x - Δx /2) sin(- Δx /2) / Δx
= - cos(1-x) * (Δx→0) lim sin(Δx /2) / (Δx/2)
= - cos(1-x)
(Δx→0) lim { ln[(x+Δx)²+1] - ln(x²+1) } / (x+Δx-x)
= (Δx→0) lim { ln[x²+2xΔx+(Δx)²+1] / (x²+1) } / Δx
= (Δx→0) lim ln {[x²+2xΔx+(Δx)²+1] / (x²+1)} ^ (1/Δx)
= (Δx→0) lim ln [1 + 2xΔx / (x²+1) + (Δx)²/(x²+1)] ^ (1/Δx)
= (Δx→0) lim ln [1 + 2xΔx / (x²+1) + (Δx)²/(x²+1)] ^ (1/Δx)
= (Δx→0) lim ln [1 + 2x/(x²+1) * Δx] ^ [(x²+1) /(2xΔx) * 2x / (x²+1)]
= ln e ^ [2x / (x²+1)]
=2x / (x²+1)
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