设y=[cos (1/x)]^3,则dy＝如题………………

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设y=[cos (1/x)]^3,则dy＝如题………………
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设y=[cos (1/x)]^3,则dy＝如题………………
设y=[cos (1/x)]^3,则dy＝
如题………………

设y=[cos (1/x)]^3,则dy＝如题………………
y'=3[cos(1/x)]^2*[cos(1/x)]'
..=3[cos(1/x)]^2*[-sin(1/x)]*(1/x)'
..=3[cos(1/x)]^2*[-sin(1/x)]*(-1/x^2)
..=3[cos(1/x)]^2[sin(1/x)]/x^2
所以dy=3[cos(1/x)]^2[sin(1/x)]/x^2 dx

结果如下：
[3/(x^2)]*[cos(1/x)]^2*sin(1/x)dx

dy=3[cos(1/x)]^2*[-sin(1/x)]*(-1)x^(-2)
=-3[cos(1/x)]^2*[-sin(1/x)]*x^(-2)

(3*cos(1/x)^2*sin(1/x)/x^2)dx

dy=3[cos(1/x)]^2*[sin(1/x)]/x^2 dx

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