主要是把根号和倒数换成指数的形式.
极限
1.
将x=2代入式子即可：
lim (x^2+4)/(x^4-2x^2+2)
x→2
=(2^2+4)/(2^4-2×2^2+2)
=0.8
2.
lim {[(1-x)^(1/2)]-3} / [2+x^(1/3)]
x→-∞
=lim {[(1-x)^(1/2)]-3}·{[(1-x)^(1/2)]+3} / [2+x^(1/3)]·{[(1-x)^(1/2)]+3}
x→-∞
=lim (-8-x) / {[2·(1-x)^(1/2)] +6 +[x^(2/3)-x^(5/3)]^(1/2) +3x^(1/3)}
x→-∞
=lim [(-8-x)/x] / {[2·(1/x^2 -1/x)^(1/2)] +6/x +[x^(2/3-2)-x^(5/3-2)]^(1/2) +3x^(1/3-1)}
x→-∞
x的最高次数不同,故不能求出定值.
导数
1.y=x^3(3x^2-2)
=3x^5-2x^3;
y'=3×5x^4 -2×3x^2
=3x^2(5x^2-2);
答案是3x^2(5x^2-2)
2.y=x^4/4+4/x^4
=x^4/4+4·x^(-4);
y'= 4x^3 /4 +4×(-4)·x^(-4-1)
=x^3-16/x^5
3.y=根号(x^3-x^2+3)
=(x^3-x^2+3)^(1/2)
y'=(1/2)×(x^3-x^2+3)^(1/2-1)×(x^3-x^2+3)'
={1/[2√(x^3-x^2+3)]}×(3x^2-2x)
=(3x^2-2x)/[2乘根号(x^3-x^2+3)]
4.y=(cos^2)x+3sin3x
y'=2[cos^(2-1)x]·(cosx)'+3(sin3x)'
=2cosx·(-sinx)+3(cos3x)·(3x)'
=-2cosx·sinx+3(cos3x)×3
=-sin2x + 9cos3x
5.y=根号[1+(ln^3)x]
=[1+(ln^3)x]^(1/2)
y'=(1/2)·[1+(ln^3)x]^(1/2 -1)·[1+(ln^3)x]'
=[1+(ln^3)x]' /{2·[1+(ln^3)x]}
= [(ln^3)x]' /{2·[1+(ln^3)x]}
=3·[ln^(3-1)x]·(lnx)' /{2·[1+(ln^3)x]}
=3·ln^2 x·(1/x) /{2·[1+(ln^3)x]}
=[3(ln^2)x]/{2x根号[1+(ln^3)x]}
6.y=x^(sin3x)
=e^[ln x^(sin3x)]
=e^[(sin3x)·ln x]
则y'=e^[(sin3x)·ln x] · [(sin3x)·ln x]'
=x^(sin3x) · [(sin3x)'·ln x + (sin3x)·(ln x)']
=x^(sin3x) · [(cos3x)·(3x)'·ln x + (sin3x)·(l/x)]
=x^(sin3x)[(3x^2)lnlnx+(x^2/linx)]
微分
1.y=4xcos4x
dy=y'·dx
=(4xcos4x)'·dx
=[(4x)'cos4x+(4x)(cos4x)']·dx
=[4cos4x+(4x)(-sin4x)·(4x)']·dx
=dy=4(cos4x-4sin4x)dx
2.y=(x^5)(e^5x)
dy=y'·dx
=[(x^5)'(e^5x) + (x^5)(e^5x)']·dx
=[5x^4·e^(5x) + (x^5)(e^5x)·(5x)']·dx
=[5x^4·e^(5x) + 5(x^5)(e^5x)]·dx
=[5(x^4)e^(5x)](1+x)dx=dy
不定积分
1.∫ x(3根号x)dx
3根号x中的3在根号的左上方
=∫ x·x^(1/3)dx
=∫ x^(1 + 1/3)dx
=∫ x^(4/3)dx
=(4/3 + 1)x^(4/3 + 1) +c
=(3/7)·x^(7/3)+c
2.∫(x+1)[(根号x)+1]dx
=∫(x+1)[x^(1/2)+1]dx
=∫[x^(3/2) +x +x^(1/2) +1]dx
=∫x^(3/2)dx +∫xdx +∫x^(1/2)dx +∫1·dx
=[1/(1+3/2)]·x^(1+3/2) + x^2/2 + =[1/(1+1/2)]·x^(1+1/2) +x +c
整理得x[(2/5)x(根号x)+(1/2)x+(2/3)(根号x)+1]+c
3.∫{[(x-1)(x+2)]/x^2}dx
=∫[(x^2 +x -2)/x^2]dx
=∫[1 +1/x -2·x^(-2)]dx
=∫1·dx +∫1/x·dx -2·∫x^(-2)·dx
=x +ln|x| -2·[1/(-2+1)]·x^(-2+1)+c
=x+ln ▏x▕ +2/x+c
4.∫(sin2x/sinx)dx
=∫(2·sinx·cosx)/sinxdx
=2·∫cosx dx
=2sinx+c