已知函数f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)求f‘(2) 导数问题
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f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)
f'(x)=(x-1)'(x-2)(x-3)(x-4)(x-5)(x-6)+(x-1)(x-2)'(x-3)(x-4)(x-5)(x-6)+(x-1)(x-2)(x-3)'(x-4)(x-5)(x-6)+(x-1)(x-2)(x-3)(x-4)'(x-5)(x-6)+(x-1)(x-2)(x-3)(x-4)(x-5)'(x-6)+(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)'
=(x-2)(x-3)(x-4)(x-5)(x-6)+(x-1)(x-3)(x-4)(x-5)(x-6)+(x-1)(x-2)(x-4)(x-5)(x-6)+(x-1)(x-2)(x-3)(x-5)(x-6)+(x-1)(x-2)(x-3)(x-4)(x-6)+(x-1)(x-2)(x-3)(x-4)(x-5)
则除了第二项,其他都有x-2
所以x=2时等于0
所以f'(2)=(2-1)(2-3)(2-4)(2-5)(2-6)=24
f'(x)=(x-1)'(x-2)(x-3)(x-4)(x-5)(x-6)+(x-1)(x-2)'(x-3)(x-4)(x-5)(x-6)+(x-1)(x-2)(x-3)'(x-4)(x-5)(x-6)+(x-1)(x-2)(x-3)(x-4)'(x-5)(x-6)+(x-1)(x-2)(x-3)(x-4)(x-5)'(x-6)+(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)'
=(x-2)(x-3)(x-4)(x-5)(x-6)+(x-1)(x-3)(x-4)(x-5)(x-6)+(x-1)(x-2)(x-4)(x-5)(x-6)+(x-1)(x-2)(x-3)(x-5)(x-6)+(x-1)(x-2)(x-3)(x-4)(x-6)+(x-1)(x-2)(x-3)(x-4)(x-5)
则除了第二项,其他都有x-2
所以x=2时等于0
所以f'(2)=(2-1)(2-3)(2-4)(2-5)(2-6)=24
2011-02-20
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对于f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)两边同时取对数,得
lnf(x)=ln(x-1)+ln(x-2)+ln(x-3)+ln(x-4)+ln(x-5)+ln(x-6)
两边同时求导得f'(x)/f(x)=1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+1/(x-5)+1/(x-6)
∴f'(x)=f(x)*[1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+1/(x-5)+1/(x-6)]
∴ 由导数形式可知,只要展开式中有(x-2)项,f'(x)那项为0
∴f'(2)=(2-1)(2-3)(2-4)(2-5)(2-6)=24
lnf(x)=ln(x-1)+ln(x-2)+ln(x-3)+ln(x-4)+ln(x-5)+ln(x-6)
两边同时求导得f'(x)/f(x)=1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+1/(x-5)+1/(x-6)
∴f'(x)=f(x)*[1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+1/(x-5)+1/(x-6)]
∴ 由导数形式可知,只要展开式中有(x-2)项,f'(x)那项为0
∴f'(2)=(2-1)(2-3)(2-4)(2-5)(2-6)=24
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记f(x)=(x-2)g(x),其中g(x)=(x-1)(x-3)(x-4)(x-5)(x-6),得g(2)=4!=24
那么f'(x)=g(x)+(x-2)g'(x)
则f'(2)=g(2)=24.
那么f'(x)=g(x)+(x-2)g'(x)
则f'(2)=g(2)=24.
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f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)
f'(x)=(x-2)(x-3)(x-4)(x-5)(x-6)+(x-1)(x-3)(x-4)(x-5)(x-6)+(x-1)(x-2)(x-4)(x-5)(x-6)+.............
f'(2)=0+1*(-1)*(-2)*(-3)*(-4)+0......+0
=24
f'(x)=(x-2)(x-3)(x-4)(x-5)(x-6)+(x-1)(x-3)(x-4)(x-5)(x-6)+(x-1)(x-2)(x-4)(x-5)(x-6)+.............
f'(2)=0+1*(-1)*(-2)*(-3)*(-4)+0......+0
=24
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f(x)=)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)
f'(x) = {(x-1)(x-3)(x-4)(x-5)(x-6) d/dx( x-2)} + {(x-2) d/dx[(x-1)(x-3)(x-4)(x-5)(x-6)]}
f'(2) = (2-1)(2-3)(2-4)(2-5)(2-6)
= 1(-1)(-2)(-3)(-4)
= 24
f'(x) = {(x-1)(x-3)(x-4)(x-5)(x-6) d/dx( x-2)} + {(x-2) d/dx[(x-1)(x-3)(x-4)(x-5)(x-6)]}
f'(2) = (2-1)(2-3)(2-4)(2-5)(2-6)
= 1(-1)(-2)(-3)(-4)
= 24
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