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所以F'(1)=1*(1-2)*(1-3)*……*(1-100)=-99!
请教已知函数f(x)=x(x-1)(x-2)(x-3)(x-4)...(x-100),求f'(1)=? f...=99!=100!用导数极限定义来解题
f(x)=(x-1)(x-2)(x-3)...(x-100),求f^-¹(1)解:可设函数g(x)=(x-2)(x-3)...(x-100),则g(1)=(-1)*(-2)*(-3)...(-99)=-99!.又函数f(x)=(x-1)g(x).求导得f'(x)=g(x)+(x-1)g'(x).∴f'(1)=g(1)=-99!.然后最重要的一步 就是 x和y互换
f(x)=(x-1)(x-2)(x-3)...(x-100),求f^-¹(1)f(x)=(x-1)(x-2)(x-3)...(x-100)=(x-1)*[(x-2)(x-3)...(x-100)]=(x-1)'*[(x-2)(x-3)...(x-100)]+(x-1)*[(x-2)(x-3)...(x-100)]'=(x-2)(x-3)...(x-100)+(x-1)*[(x-2)(x-3)...(x-100)]'f'(1)=(-1)*(-2)*...*(-99)=-...
若f(x)=x(x-1)(x-2)(x-3)...(x-100),则f(x)在x=0处的导数是多少?f'(2)可解释为在x=2处的切线斜率 所以k=f'(2)=3 f'(0)=-1-2-3.-100=-(1+100)*100/2=-5050 因为f'(x)=x'(x-1).(x-100)+x(x-1)'(x-2)...(x-100)+x(x-1)(x-2)'.因此从第2项起都有x将0带入为0
已知函数f(x)=x(x-1)(x-2)(x-3)(x-4)...(x-100)求f'(0)=f'(x)=x'*(x-1)(x-2)(x-3)(x-4)...(x-100)+x(x-1)'*(x-2)(x-3)(x-4)...(x-100)+……+x(x-1)(x-2)(x-3)(x-4)...(x-100)'=(x-1)(x-2)(x-3)(x-4)...(x-100)+x(x-2)(x-3)(x-4)...(x-100)+……+x(x-1)(x-2)(x-3)(x-4)....
设f(X)=(X-1)(X-2)(X-3)...(X-101),求f′(1)求导公式 f(x)=m(x)*n(x)则 f'(x) =m'(x)*n(x)+m(x)*n'(x)将f(x)=(x-1)(x-2)...(x-101)记作f(x)=(x-1)*g(x)其中g(x)=(x-2)...(x-101)f'(x)=(x-1)*g'(x)+g(x)f'(1)=g(1) (不用计算g'(x)!)=(-1)×(-2)×(-3)×...×...
已知:f(x)=(x-1)(x-2)(x-3)……(x-100)求f'(1)解:可设函数g(x)=(x-2)(x-3)...(x-100),则g(1)=(-1)*(-2)*(-3)...(-99)=-99!.又函数f(x)=(x-1)g(x).求导得f'(x)=g(x)+(x-1)g'(x).∴f'(1)=g(1)=-99!.
函数f(x)=x(x-1)(x-2)(x-3)...(x-100)在x=0处的导数值为?紧急!是0啊,把前两项看成一项,后面看成一项,则先对x(x-1)求导得2x,后面的就不用问了,因为x=0.
已知f(x)=x(x-1)(x-2)(x-3)……(x-100),则f'(x)=对数求导,还有f(x)为复合函数,lnf(x)求导时为1/f(x)乘以f'(x)设F(x)=x(x-1)(x-2)(x-3)···(x-100)。F'(x)=(x-1)(x-2)(x-3)···(x-100)+x(x-2)(x-3)···(x-100)+x(x-1)(x-3)···(x-100)+x(x-1)(x-2)(x-3)···(x-49)(x-51)*...