设z=(x/y)sin(x²y³)，求∂z/∂x，∂z/∂y．

设z=(x/y)sin(x²y³)，求∂z/∂x，∂z/∂y．

解：利用积的求导法则，得 ∂z/∂x=[∂/∂x(x/y)]•sin(x²y³)+x/y•∂/∂x[sin(x²y³)] =1•sin(x²y³)+x/y•cos(x²y³)•2xy³ =sin(x²y³)+2x²y²cos(x²y³)， ∂z/∂y=[/y(x/y)]•sin(x²y³)+x/y•∂/∂y[sin(x²y³)] =-(x/y²)sin(x²y³)+x/y• cos(x²y³)•3x²y³ =x/y²sin(x²y³)+3x³ycos(x²y³)