作业帮设函数f(x)=[3/(5-x)]+(x²/5),则f'(0)=
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设函数f(x)=[3/(5-x)]+(x²/5),则f'(0)=
设函数f(x)=[3/(5-x)]+(x²/5),则f'(0)=
设函数f(x)=[3/(5-x)]+(x²/5),则f'(0)=
f'=[(0-3(5-x)')]/(5-x)^2+2x/5
=3/(5-x)^2+2x/5
所以f'(0)=3/25
v=1/(5-x) v'=(-1)/(5-2x)2
f'(x)=3v'*(-1)+1/5*2x
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