Question

求隐函数的二阶导数

求隐函数b^2x^2+ a^2y^2=a^2b^2的二阶导数，要详细步骤。我算答案总是不对，正确答案为：-b^4/a^2y^3

Answer 1

b²x²+a²y²=a²b²
d(b²x²+a²y²)=d(a²b²)
2b²xdx+2a²ydy=0
dy/dx=(-b²/a²)(x/y)
d²y/dx²=(-b²/a²)[(y-xy')/y²]
=(-b²/a²)[y+(b²x²)/(a²y)]/y²
=(-b²/a²)(a²y²+b²x²)/(a²y³) [题目中已经给出b²x²+a²y²=a²b²]
=(-b²/a²)(a²b²)/(a²y²)
=-b^4/(a²y³)

Answer 2

解:f(sinx)=3-cos(2x)=3-(1-2sinx*sinx)=2+2sinx*sinx故f(x)=2+2*x*x故f(cosx)=2+2cosx*cosx=3+2cosx*cosx-1=3+cos2x