【基础高数题1、设函数f（x）=limt趋向于无限大{t^2*sin(x/t)*[φ（x+π/t)-φ(x)]}其中φ具有二阶导数,求f'(x).2、设方程x=(3t^2)+2t+3(e^ysint)-y+1=0,求d2y/dx2t=0】
基础高数题
1、设函数f（x）=limt趋向于无限大{t^2*sin(x/t)*[φ（x+π/t)-φ(x)]}
其中φ具有二阶导数,求f'(x).
2、设方程x=(3t^2)+2t+3(e^ysint)-y+1=0,求d2y/dx2t=0
1.f(x)=lim{t^2*sin(x/t)*[φ（x+π/t)-φ(x)]}
=limt^2*sin(x/t)[φ'(x)*(π/t)+(1/2)*φ''(x)*(π/t)^2+o(π/t)^2](泰勒公式)
=limt^2*sin(x/t)*φ'(x)*(π/t)
=limπx*[sin(x/t)/(x/t)]φ'(x)(x/t→0)
=πxφ'(x)
∴f'(x)=πφ'(x)+πxφ''(x)
2.dx/dt=6t+2
式两端对t求导得e^y(dy/dt)sint+e^ycost-dy/dt=0
即dy/dt=e^ycost/(1-e^ysint)
∴dy/dx=(dy/dt)/(dx/dt)=e^ycost/(1-e^ysint)(6t+2)
同样d2y/dx2=[d(dy/dx)/dt]/(dx/dt)可求其表达式
当t=0有x=3,y=1
计算可得d2y/dx2=(4e^2-6e)/8=e(2e-3)/4