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设函数y=x^x+ln(arctan5x),求其导数dy/dx、微分dy
人气:273 ℃ 时间:2019-08-17 21:18:39
解答
y'=(x^x)'+(ln(arctan5x)'
设f(x)=x^x
lnf(x)=xlnx
1/f(x)f'(x)=lnx+1
f'(x)=f(x)(lnx+1)=x^x(lnx+1)
ln(arctan5x)'=1/(arctan5x) * (1/(1+25x^2))*5=5/[(1+25x^2)(arctan5x)]
dy/dx=x^x(lnx+1)+5/[(1+25x^2)(arctan5x)]
dy={x^x(lnx+1)+5/[(1+25x^2)(arctan5x)]}dx
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