求cos²(1/x)的导数和微分 y=cos（x+y）的导数怎么求？
cos^2(1/x)\u7684\u5bfc\u6570\u600e\u4e48\u6c42
\u81ea\u5916\u5230\u5185\uff0c\u94fe\u5f0f\u6cd5\u5219\u3002
y' = -sin ( x + y )/1 + sin ( x + y ) \u3002
\u5206\u6790\u8fc7\u7a0b\u5982\u4e0b\uff1a
y = cos ( x+ y)
y' = [ cos ( x + y )]' * ( x + y)' \u94fe\u5f0f\u6cd5\u5219\u3002
y' = -sin ( x + y ) * ( 1 + y') \u51fd\u6570\u6c42\u5bfc\u6cd5\u5219\uff0ccos ( x+y)\u7684\u5bfc\u6570\u662f-sin(x+y)\uff0c\u540e\u9762\u62ec\u53f7\u91cc\u9762x\u7684\u5bfc\u6570\u662f1\uff0cy\u7684\u5bfc\u6570\u6211\u4eec\u73b0\u5728\u8fd8\u4e0d\u77e5\u9053\uff08\u6b63\u662f\u6211\u4eec\u8981\u6c42 \u7684\uff09\uff0c\u6240\u4ee5\u7528y'\u8868\u793a\u3002
y' = -sin ( x + y ) + y' * [-sin (x + y)]
y' + y'sin ( x + y ) = -sin ( x + y )
y' * [ 1 + sin ( x + y \uff09] = -sin ( x + y )
y' = -sin ( x + y )/1 + sin ( x + y )
\u6269\u5c55\u8d44\u6599\uff1a
\u94fe\u5f0f\u6cd5\u5219\u662f\u5fae\u79ef\u5206\u4e2d\u7684\u6c42\u5bfc\u6cd5\u5219\uff0c\u7528\u4ee5\u6c42\u4e00\u4e2a\u590d\u5408\u51fd\u6570\u7684\u5bfc\u6570\u3002\u6240\u8c13\u7684\u590d\u5408\u51fd\u6570\uff0c\u662f\u6307\u4ee5\u4e00\u4e2a\u51fd\u6570\u4f5c\u4e3a\u53e6\u4e00\u4e2a\u51fd\u6570\u7684\u81ea\u53d8\u91cf\u3002\u5982\u8bbef(x)=3x\uff0cg(x)=3x+3\uff0cg(f(x))\u5c31\u662f\u4e00\u4e2a\u590d\u5408\u51fd\u6570\uff0c\u5e76\u4e14g\u2032(f(x))=9
\u94fe\u5f0f\u6cd5\u5219\uff0c\u82e5h(a)=f[g(x)]\uff0c\u5219h'(a)=f\u2019[g(x)]g\u2019(x)\u3002
\u94fe\u5f0f\u6cd5\u5219\u7528\u6587\u5b57\u63cf\u8ff0\uff0c\u5c31\u662f\u201c\u7531\u4e24\u4e2a\u51fd\u6570\u51d1\u8d77\u6765\u7684\u590d\u5408\u51fd\u6570\uff0c\u5176\u5bfc\u6570\u7b49\u4e8e\u91cc\u51fd\u6570\u4ee3\u5165\u5916\u51fd\u6570\u7684\u503c\u4e4b\u5bfc\u6570\uff0c\u4e58\u4ee5\u91cc\u8fb9\u51fd\u6570\u7684\u5bfc\u6570\u3002\u201d
\u5e38\u7528\u5bfc\u6570\u516c\u5f0f\uff1a
1.y=c(c\u4e3a\u5e38\u6570) y'=0
2.y=x^n y'=nx^(n-1)
3.y=a^x y'=a^xlna\uff0cy=e^x y'=e^x
4.y=logax y'=logae/x\uff0cy=lnx y'=1/x
5.y=sinx y'=cosx
6.y=cosx y'=-sinx
7.y=tanx y'=1/cos^2x
8.y=cotx y'=-1/sin^2x
9.y=arcsinx y'=1/\u221a1-x^2
10.y=arccosx y'=-1/\u221a1-x^2
dy/dx
= 2cos (1/x) .d/dx cos(1/x)
= 2cos (1/x) .[-sin(1/x)] . d/dx(1/x)
= (2/x^2).cos (1/x) .sin(1/x)
= sin(2/x)/x^2
dy =[sin(2/x)/x^2] dx
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