高等数学求dy
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- 发布时间:2026-04-28 18:08:57
lny=lnx/sin(1/x)
y'/y=[sin(1/x)/x-lnxcos(1/x)*(-1/x^2)]/sin^2(1/x)
y'=y*[sin(1/x)/x-lnxcos(1/x)*(-1/x^2)]/sin^2(1/x)
lny=lnx/sin(1/x)
y'/y=[sin(1/x)/x-lnxcos(1/x)*(-1/x^2)]/sin^2(1/x)
y'=y*[sin(1/x)/x-lnxcos(1/x)*(-1/x^2)]/sin^2(1/x)