作业帮y = x^2 sin 4x,求dy,求dy
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y = x^2 sin 4x,求dy,求dy
y = x^2 sin 4x,求dy,
求dy
y = x^2 sin 4x,求dy,求dy
y'=2xsin4x-x²cos4x·4
所以
dy=(2xsin4x-4x²cos4x)dx
y=ln√4＋t²=1/2ln(4+t²)
y'=1/2·1/(4+t²)·2t
=t/(4+t²)
dy=t/(4+t²)dt
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