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ln (tanx )二次求导
人气:231 ℃ 时间:2020-03-08 03:41:20
解答
设f(x)=ln(tanx)

f'(x)=1/tanx *(tanx)'
=1/tanx *1/cos^2 x
=cosx/sinx *1/cos^2 x
=1/(sinxcosx)
=2/sin2x
f''(x) =-2 *2*cos2x/sin^2 2x
=-4cos2x/sin^2 2x
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