作业帮解∫sinx^2cosx^5dx,y=(sin1/x+1/x),x→0时的极限,lim(1+sinx)^1/x,x→0,lim(x^2-4)/(x^2+2x-8),x→2;求y=ln[(1-sinx)/(1+sinx)]的导数.最后一题要求详解~~
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/04 09:43:39
![解∫sinx^2cosx^5dx,y=(sin1/x+1/x),x→0时的极限,lim(1+sinx)^1/x,x→0,lim(x^2-4)/(x^2+2x-8),x→2;求y=ln[(1-sinx)/(1+sinx)]的导数.最后一题要求详解~~](/uploads/image/z/1747572-60-2.jpg?t=%E8%A7%A3%E2%88%ABsinx%5E2cosx%5E5dx%2Cy%3D%28sin1%2Fx%2B1%2Fx%29%2Cx%E2%86%920%E6%97%B6%E7%9A%84%E6%9E%81%E9%99%90%2Clim%281%2Bsinx%29%5E1%2Fx%2Cx%E2%86%920%2Clim%28x%5E2-4%29%2F%28x%5E2%2B2x-8%29%2Cx%E2%86%922%3B%E6%B1%82y%3Dln%5B%281-sinx%29%2F%281%2Bsinx%29%5D%E7%9A%84%E5%AF%BC%E6%95%B0.%E6%9C%80%E5%90%8E%E4%B8%80%E9%A2%98%E8%A6%81%E6%B1%82%E8%AF%A6%E8%A7%A3%7E%7E)
解∫sinx^2cosx^5dx,y=(sin1/x+1/x),x→0时的极限,lim(1+sinx)^1/x,x→0,lim(x^2-4)/(x^2+2x-8),x→2;求y=ln[(1-sinx)/(1+sinx)]的导数.最后一题要求详解~~
解∫sinx^2cosx^5dx,y=(sin1/x+1/x),x→0时的极限,lim(1+sinx)^1/x,x→0,lim(x^2-4)/(x^2+2x-8),x→2;
求y=ln[(1-sinx)/(1+sinx)]的导数.最后一题要求详解~~
解∫sinx^2cosx^5dx,y=(sin1/x+1/x),x→0时的极限,lim(1+sinx)^1/x,x→0,lim(x^2-4)/(x^2+2x-8),x→2;求y=ln[(1-sinx)/(1+sinx)]的导数.最后一题要求详解~~
∫(sinx)^2(cosx)^5dx
=∫(sinx)^2(1-(sinx)^2)cosxdx
=∫(sinx)^2dsinx-∫(sinx)^4dsinx
=(sinx)^3/3 -(sinx)^5/5 +C
lim(x→0)(sin1/x+1/x)
x→0 1/x→∞
lim(x→0)sin(1/x)不存在
lim(x→0) (1+sinx)^(1/x) =lim( x→0)[(1+sinx)^(1/sinx)]^(sinx/x)
lim(x→0)sinx/x=1
=lim(sinx→0,x→0)(1+sinx)^(1/sinx)^(sinx/x)
=e
lim(x→2)(x^2-4)/(x^2+2x-8)
=lim(x→2)(x^2-4)'/(x^2+2x-8)'
=lim(x→2)(2x)/(2x+2)
=2/3
y=ln(1-sinx)/(1+sinx)
(1-sinx)/(1+sinx)=2/(1+sinx)-1 [(1-sinx)/(1+sinx)]'= -2(sinx)'/(1+sinx)^2= -2cosx/(1+sinx)^2
y'=[(1-sinx)/(1+sinx)]' / [(1-sinx)/(1+sinx)]
=[ - cosx/(1+sinx)^2 ] *[(1+sinx)/(1-sinx)]
= -2cosx/[(1+sinx)(1-sinx)]
=-2cosx/(cosx)^2= -2/cosx
ytuty
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