题目内容
已知f(x)与g(x)均可异,且f(x)>0,设y=f(x)g(x),求y′.
解:y=f(x)g(x)=e g(x)lnf(x),则eg(x)lnf(x)可看作由eu与u=g(x)lnf(x)复合而成.
∴y′=eg(x)lnf(x)[g(x)lnf(x)]′
=eg(x)lnf(x)[g′(x)lnf(x)+g(x)·f′(x)]
=f(x)g(x)[g(x)lnf(x)+g(x) ].
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题目内容
已知f(x)与g(x)均可异,且f(x)>0,设y=f(x)g(x),求y′.
解:y=f(x)g(x)=e g(x)lnf(x),则eg(x)lnf(x)可看作由eu与u=g(x)lnf(x)复合而成.
∴y′=eg(x)lnf(x)[g(x)lnf(x)]′
=eg(x)lnf(x)[g′(x)lnf(x)+g(x)·f′(x)]
=f(x)g(x)[g(x)lnf(x)+g(x) ].