企源知识库

设函数ƒ(x)=sine^-x，求ƒ(0)+ƒ´(0)+ƒ´´(0)．
【正确答案】：因ƒ(0)=sinl，ƒ'(x)=-e^-xcose^-x，ƒ′(0)=-cosl．ƒ”(x)=e^-x(cose^-x-e^-xsine^-x)，ƒ”(0)=cosl-sinl．故ƒ(0)+ƒ'(0)+ƒ”(0)=0．