Question

一道高等数学，偏微分，过程。

Answer 1

w=sinm*e^n
∂w/∂x = cosm(∂m/∂u)(∂u/∂x)e^n + sinm*e^n(∂n/∂v)(∂v∂x)
= cosm(a^ulna)e^n + sinm*e^n(-sinv)y = (lna)cosm*e^n*a^u - ysinm*e^n*sinv,

∂w/∂y = cosm(∂m/∂u)(∂u/∂y)e^n + sinm*e^n(∂n/∂v)(∂v∂y)
= cosm(a^ulna)e^n + sinm*e^n(-sinv)x = (lna)cosm*e^n*a^u - xsinm*e^n*sinv

∂^2w/∂y∂x = lna[-sinm(∂m/∂u)(∂u/∂x)e^n*a^u+cosm*a^u*e^n(∂n/∂v)(∂v∂x)
+cosm*e^n*a^u*lna(∂u/∂x)] - sinm*e^n*sinv-x[cosm(∂m/∂u)(∂u/∂x)e^n*sinv
+sinm*sinv*e^n(∂n/∂v)(∂v∂x)+sinm*e^n*cosv(∂v∂x)]
=lna[-sinm*(a^ulna)e^n*a^u+cosm*a^u*e^n(-sinv)y+cosm*e^n*a^u*lna]
- sinm*e^n*sinv-x[cosm(a^ulna)e^n*sinv+sinm*sinv*e^n(-sinv)y+sinm*e^n*cosv*y]
=-(lna)^2*sinm*a^(2u)e^n-(lna)(x+y)cosm*a^u*e^n*sinv+(lna)^2cosm*e^n*a*u
- sinm*e^n*sinv+xysinm*e^n(sinv)^2-xysinm*e*n*cosv.

∂^2w/∂y^2 = lna[-sinm(∂m/∂u)(∂u/∂y)e^n*a^u+cosm*a^u*e^n(∂n/∂v)(∂v∂y)
+cosm*e^n*a^u*lna(∂u/∂y)] - sinm*e^n*sinv-x[cosm(∂m/∂u)(∂u/∂y)e^n*sinv
+sinm*sinv*e^n(∂n/∂v)(∂v∂y)+sinm*e^n*cosv(∂v∂y)]
=lna[-sinm*(a^ulna)e^n*a^u+cosm*a^u*e^n(-sinv)x+cosm*e^n*a^u*lna]
- sinm*e^n*sinv-x[cosm(a^ulna)e^n*sinv+sinm*sinv*e^n(-sinv)x+sinm*e^n*cosv*x]
=-(lna)^2*sinm*a^(2u)e^n-2(lna)xcosm*a^u*e^n*sinv+(lna)^2cosm*e^n*a*u
- sinm*e^n*sinv+x^2*sinm*e^n(sinv)^2-x^2*sinm*e*n*cosv.