y=(x-1)(x-2)(x-3)……(x-2005) 求导两边同时取自然对数得:lny=ln(x-1)+ln(x-2)+ln(x-3)+……+ln(x-2005)两边同时对x求导得:y '/y=1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)所以y '=y[1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)]=(x-1

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y=(x-1)(x-2)(x-3)……(x-2005) 求导两边同时取自然对数得:lny=ln(x-1)+ln(x-2)+ln(x-3)+……+ln(x-2005)两边同时对x求导得:y '/y=1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)所以y '=y[1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)]=(x-1

y=(x-1)(x-2)(x-3)……(x-2005) 求导两边同时取自然对数得:lny=ln(x-1)+ln(x-2)+ln(x-3)+……+ln(x-2005)两边同时对x求导得:y '/y=1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)所以y '=y[1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)]=(x-1
y=(x-1)(x-2)(x-3)……(x-2005) 求导
两边同时取自然对数得:
lny=ln(x-1)+ln(x-2)+ln(x-3)+……+ln(x-2005)
两边同时对x求导得:
y '/y=1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)
所以y '=y[1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)]
=(x-1)(x-2)(x-3)……(x-2005)[1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)]
为什么需要 y的导数比上y?

y=(x-1)(x-2)(x-3)……(x-2005) 求导两边同时取自然对数得:lny=ln(x-1)+ln(x-2)+ln(x-3)+……+ln(x-2005)两边同时对x求导得:y '/y=1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)所以y '=y[1/(x-1)+1/(x-2)+1/(x-3)+……+1/(x-2005)]=(x-1
因为左边lny的导数等于(lny)'乘以(y)'等于1/y乘以y'等于y'/y,看你理解啦