2. 考研
  3. [2015年] (I)设函数u(x)，v(x)可导，利用导数定义证明[u(x)v(x)]’=u’(x)v(x)+u(x)v’(x)； (Ⅱ)设函数u1(x)，u2(x)，…，un(x)可导，f(x)=u1(x)u2(x)…un(x)，写出f(x)的求导公式

[2015年] (I)设函数u(x)，v(x)可导，利用导数定义证明[u(x)v(x)]’=u’(x)v(x)+u(x)v’(x)； (Ⅱ)设函数u1(x)，u2(x)，…，un(x)可导，f(x)=u1(x)u2(x)…un(x)，写出f(x)的求导公式

admin2019-04-08  32
问题 [2015年] (I)设函数u(x)，v(x)可导，利用导数定义证明[u(x)v(x)]’=u’(x)v(x)+u(x)v’(x)；
(Ⅱ)设函数u1(x)，u2(x)，…，un(x)可导，f(x)=u1(x)u2(x)…un(x)，写出f(x)的求导公式．
选项
答案(I)令f(x)=u(x)v(x)，由 △f=f(x+△x)-f(x)=u(x+△x)v(x+△x)一u(x)v(x) =u(x+△x)v(x+△x)一u(x)v(x+△)+u(x)v(x+△)一u(x)v(x) =v(x+△)[u(x+△)一u(x)]+u(x)[v(x+△)-v(x)] =v(x+△)△u+u(x)△v， 得到 [*] (Ⅱ)由上题中f(x)的导数公式的形式易得到 f’(x)=[u1(x)u2(x)…un(x)]’一{[u1(x)]u2(x)…un(x)}’ =[u1(x)]’[u2(x)…un(x)]+u1(x)[u2(x)…un(x)]’ =u’1(x)u2(x)…un(x)+u1(x){[u2(x)]u3(x)…un(x)}’ =u’1(x)u2(x)…un(x)+u1(x)u’2(x)…un(x)+u1(x)u2(x)[u3(x)…un(x)]’ =… =u’1(x)u2(x)…un(x)+u1(x)u’2(x)…un(x)+…+u1(x)u2(x)…u’n(x)．
解析
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考研数学一

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