f（x）=sin（2^x-2^x\1)的導(dǎo)數(shù)
f（x）=sin（2^x-2^x\1)的導(dǎo)數(shù)
數(shù)學(xué)人氣：858 ℃時間：2019-08-26 06:43:32
優(yōu)質(zhì)解答
f（x）=sin[2^x-2^(1/x)]f'(x)=cos[2^x-2^(1/x)]* [2^x-2^(1/x)]'=[2^x*ln2-2^(1/x)*ln2*(1/x)']*cos[2^x-2^(1/x)]=[2^x+1/x²*2^(1/x)]ln2*cos[2^x-2^(1/x)]
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