对f(x)求导,附过程
f(x)=x+2√(x²-3x+3)
f'(x)=[x+2√(x²-3x+3)]'
=x'+2[√(x²-3x+3)]'
=1+2(x²-3x+3)'/[2√(x²-3x+3)]
=1+(2x-3)/√(x²-3x+3)
用到的公式:
(x^n)'=nx^(n-1)
[f(u)]'=u'f'(u)
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f(x)=x+2√(x²-3x+3)
f'(x)=[x+2√(x²-3x+3)]'
=x'+2[√(x²-3x+3)]'
=1+2(x²-3x+3)'/[2√(x²-3x+3)]
=1+(2x-3)/√(x²-3x+3)
用到的公式:
(x^n)'=nx^(n-1)
[f(u)]'=u'f'(u)