Question

求全微分z=(x^2+y^2)/(x^2-y^2)

Answer 1

全微分
dz=∂z/∂x *dx + ∂z/∂y *dy
在这里z=(x²+y²)/(x²-y²)
那么
∂z/∂x=[∂(x²+y²)/∂x *(x²-y²) -(x²+y²)*∂(x²-y²)/∂x] /(x²-y²)²
=[2x*(x²-y²) -(x²+y²)*2x] / (x²-y²)²
= -4xy² /(x²-y²)²
同理
∂z/∂y=[∂(x²+y²)/∂y *(x²-y²) -(x²+y²)*∂(x²-y²)/∂y] /(x²-y²)²
=[2y*(x²-y²) +(x²+y²)*2y] / (x²-y²)²
= 4x²y / (x²-y²)²
所以
dz= -4xy² /(x²-y²)² dx + 4x²y / (x²-y²)² dy