f0(x)=sinx,f1(x)=f'0(x),f2(x)=f'1(x),...,fn+1(x)=f'n(x) (n∈N)由此归纳推测f2009(x)
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f0(x)=sinx,f1(x)=f''0(x),f2(x)=f''1(x),...,fn+1(x)=f''n(x)(n∈N)由此归纳推测f2009(x)f0(x)=sinx,f1(x)=f''0(x),f2
f0(x)=sinx,f1(x)=f'0(x),f2(x)=f'1(x),...,fn+1(x)=f'n(x) (n∈N)由此归纳推测f2009(x)
f0(x)=sinx,f1(x)=f'0(x),f2(x)=f'1(x),...,fn+1(x)=f'n(x) (n∈N)由此归纳推测f2009(x)
f0(x)=sinx,f1(x)=f'0(x),f2(x)=f'1(x),...,fn+1(x)=f'n(x) (n∈N)由此归纳推测f2009(x)
f1(x)=(sinx)'=cosx
f2(x)=(cosx)'=-sinx
f3(x)=(-sinx)'=-(sinx)'=-cosx
f4(x)=(-cosx)'=-(cosx)'=sinx=f0(x)
所以是4个一循环
2009÷4余数是1
所以f2009(x)=f1(x)=cosx
f0(x)=xe^x,f1(x)=f0'(x),f2(x)=f1'(x),.,fn(x)=f'n-1(x)(n∈N^*),f2012(0)=?
f0(x)=sinx,f1(x)=f'0(x),f2(x)=f'1(x),...,fn+1(x)=f'n(x) (n∈N)由此归纳推测f2009(x)
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