求极限导数微分不定积分求极限1.lim (x^2+4)/(x^4-2x^2+2)x→22.lim 分子是[根号(1-x)
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4个回答

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

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