y=(x-1)/(x+1)=1 - 2/(x+1)
y'=2/(x+1)^2
y=(1-√x)(1+√x)=1-x
y'=-1
y=(x根号x/三次根号x)cosx
=x^(7/6)*cosx
y'=(7/6)x^(1/6)*cosx - x^(7/6)*sinx
y=xsinx+√x = xsinx + x^(1/2)
y'=sinx+x*cosx + 1/(2√x)
y=Inx/x+1-2^x
y'=[(1/x)*(x+1)-lnx]/(x+1)^2 -2^x*ln2
=(1/x - lnx +1)/(x+1)^2 - 2^x*ln2
最后一题有歧义,给两种方式吧.
y=lnx/x + 1 - 2^x
y'=(1/x * x - lnx)/x^2 - 2^x*ln2
=(1-lnx)/x^2 - 2^x*ln2