设y=cos√x+2∧x,求dy
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设y=cos√x+2∧x,求dy
设y=cos√x+2∧x,求dy
设y=cos√x+2∧x,求dy
y=cos√x+2^x
y'=-sin√x/(2√x)+ln2*2^x
dy=[-sin√x/(2√x)+ln2*2^x]dx
dy=(-sin√x/√x+ln2*2^x)dx
dy=[(-sin√x)/(2√x)+2x]dx.
dy也就是求导,即dy=-sin√x{1/2(x^(-1/2))}+2^x*ln2.其实式子很简单,怕你看不懂
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