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数学
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求y=ln(sin2x+2^x)的导数
人气:349 ℃ 时间:2020-01-24 02:34:03
解答
y=ln(sin2x+2^x)
y'=1/(sin2x+2^x) * (sin2x + 2^x)'
=1/(sin2x+2^x) * (cos2x * (2x)' + ln2 * 2^x)
=(2cos2x + ln2 * 2^x) / (sin2x+2^x)
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