对函数y=ln[cos(arctan(sinx))]求导
最后答案是-sinxcosx/(1+sinx^2)
人气:350 ℃ 时间:2020-02-01 21:22:58
解答
y=f{g[h(p(x))]}y'=f'(g)g'(h)h'(p)p'(x)y'=1/cos(arctan(sinx)) *(-sin(arctan(sinx))*cosx/(1+sinx^2)= -tan(arctan(sinx))*cosx/(1+sinx^2)=-sinxcosx/(1+sinx^2)
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