求导arctany/xIn√(x²+y²)(x²+y²)si

网友回答

全微分吗?z=arctan(y/x)
∂z/∂x=1/(1+y²/x²)*y=x²y/(x²+y²)
∂z/∂y=1/(1+y²/x²)*1/x=x/(x²+y²)
dz=x²y/(x²+y²)dx+x/(x²+y²)dy
z=ln√(x²+y²)
∂z/∂x=1/√(x²+y²)*1/2√(x²+y²)*2x=x/(x²+y²)
∂z/∂y=1/√(x²+y²)*1/2√(x²+y²)*2y=y/(x²+y²)
dz=x/(x²+y²)dx+y/(x²+y²)dy
z=(x²+y²)sin[3/(x²+y²)]
∂z/∂x=2xsin[3/(x²+y²)]+(x²+y²)*cos[3/(x²+y²)]*[-6x/(x²+y²)²]=2xsin[3/(x²+y²)]-6xcos[3/(x²+y²)]*/(x²+y²)
∂z/∂y=2ysin[3/(x²+y²)]+(x²+y²)*cos[3/(x²+y²)]*[-6y/(x²+y²)²]=2ysin[3/(x²+y²)]-6ycos[3/(x²+y²)]*/(x²+y²)
dz=2xsin[3/(x²+y²)]-6xcos[3/(x²+y²)]*/(x²+y²)dx+2ysin[3/(x²+y²)]-6ycos[3/(x²+y²)]*/(x²+y²)dy
======以下答案可供参考======
供参考答案1:
太难了。。。。。高数都忘记了