设函数ƒ(x)=sine-x，求ƒ(0)+ƒ´(0)+ƒ´&ac

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．