mutangbuding
幼苗
共回答了18个问题采纳率:94.4% 举报
(1)∂z/∂x=4x³-8xy²,∂z/∂y=4y³-8x²y
(2)y'=(1+x²)'/(1+x²)=2x/(1+x²),dy=y'dx=[2x/(1+x²)]dx;
(3)定义域 (0,∞);x→0,f(x)→+∞,x→+∞,f(x)→+∞;令 f'(x)=4x-(1/x)=0,求得函数驻点x= 1/2;
极小值:f(1/2)=2(1/2)²-ln(1/2)=(1/2)+ln2;当 01/2,单调递增;
(4)x→0,lim{[1/ln(1+x)]-(1/x)}=lim{[x-ln(1+x)]/[xln(1+x)]}=lim{[1- 1/(1+x)]/[x/(1+x) + ln(1+x)]}
=lim{x/[x+(1+x)ln(1+x)]}=lim{1/[1+1+ln(1+x)]}=1/(2+0)=1/2;
(5)dx+xydy=y³dx+ydy → y(x-1)dy=(y³-1)dx → ydy/(y³-1)=dx/(x-1);
∫ydy/(y³-1)=∫dx/(x-1) → (1/2)∫[1/(y-1) - (y-1)/(y²+y+1)]dy=ln(x-1) →
ln(y-1)-∫[(y+ 1/2 - 3/2)/(y²+y+1)]dy=2ln(x-1)
ln(y-1)-(1/2)*ln(y²+y+1)-(3/2)∫[1/(y²+y+1)]dy=2ln(x-1)
ln(y-1)-ln√(y²+y+1)-(3/2)∫dy/[(y+ 1/2)²+ (3/4)]=2ln(x-1)
ln(y-1)-ln√(y²+y+1)-∫d(2y+1)/[3(2y+1)²+1)]=2ln(x-1)
ln(y-1)-ln√(y²+y+1)-(√3/3)artan[(2y+1)√3]+C=2ln(x-1)
1年前
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