f(x)=(x-1)e^x-1,则f'(x)=x-1为e的指数
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f(x)=(x-1)e^x-1,则f'(x)=x-1为e的指数
f(x)=(x-1)e^x-1,则f'(x)=
x-1为e的指数
f(x)=(x-1)e^x-1,则f'(x)=x-1为e的指数
f(x)=(x-1)e^(x-1)
f'(x)=e^(x-1)+(x-1)e^(x-1)
=xe^(x-1)
f'(x)=(x-1)'*e^(x-1)+(x-1)*[e^(x-1)]'
=e^(x-1)+(x-1)e^(x-1)
=xe^(x-1)
解
f'(x)=[(x-1)e^(x-1)]'
=(x-1)'e^(x-1)+(x-1)[e^(x-1)]'
=e^(x-1)+(x-1)e^(x-1)(x-1)'
=e^(x-1)+(x-1)e^(x-1)
=xe^(x-1)
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