設f(x)=x√(x^2+4)，求f(x)在x=0的導數?
回答 (2)
2016-05-15 5:57 am
f(x)=x√(x^2+4)
f'(x)=0.5(x^2+4)^-0.5 (2x)x+√(x^2+4)
f'(x)=(x^2)(x^2+4)^-0.5+√(x^2+4)
f'(0)=√(0^2+4)
f'(0)=2
f'(x)=0.5(x^2+4)^-0.5 (2x)x+√(x^2+4)
f'(x)=(x^2)(x^2+4)^-0.5+√(x^2+4)
f'(0)=√(0^2+4)
f'(0)=2
2016-05-10 11:08 am
Sol
f(0)=0
lim(x->0)_[f(x)-f(0)]/(x-0)
=lim(x->0)_x√(x^2+4)/x
=lim(x->0)_√(x^2+4)
=2
f(0)=0
lim(x->0)_[f(x)-f(0)]/(x-0)
=lim(x->0)_x√(x^2+4)/x
=lim(x->0)_√(x^2+4)
=2
收錄日期: 2021-04-18 14:48:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160509175113AAmtZDg