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y = (sin 5x)^ln x 求导
人气:351 ℃ 时间:2020-01-20 15:47:40
解答
y' = (a^x)' = a^xlnxy' = (x^a)' = ax^(a-1)y=(sin5x)^(lnx)y' = [(sin5x)^lnx]·ln(lnx)·(1/x) + lnx(sin5x)^(lnx - 1)·5cos5x = [(sin5x)^lnx]]·ln(lnx)·(1/x) + 5(cos5x)(lnx)(sin5x)^(lnx - 1)
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