Question

If z=(x+y)ey and x=u2+v2 and y=u2−v2, find the following partial derivatives using the chain rule. Enter your answers as functions of u and v.
∂z/∂u= ______
∂z/∂v= ______

Answer 1

Answer:

Step-by-step explanation:

Given the functions  z=(x+y)$e^y$ and x=u²+v² and y=u²−v²

Using the composite derivative formula;

∂z/∂u= ∂z/∂x*∂x/∂u+∂z/∂y*∂y/∂u

∂z/∂u = y$e^y$*2u + [(x+y)$e^y$+x$e^y$]*2u

∂z/∂u =y$e^y$*2u + 2u[x$e^y$+y$e^y$+x$e^y$]

∂z/∂u = y$e^y$*2u + 2u[2x$e^y$+y$e^y$]

∂z/∂u = 2u[u²−v²]$e^{u^2-v^2}$+ 2u[2(u²+v²)$e^{u^2-v^2}$+y$e^{u^2-v^2}$]]

∂z/∂v= ∂z/∂x*∂x/∂v+∂z/∂y*∂y/∂v

∂z/∂v = y$e^y$*2v + [(x+y)$e^y$+x$e^y$]*-2v

∂z/∂v =y$e^y$*2v -2v[x$e^y$+y$e^y$+x$e^y$]

∂z/∂v = y$e^y$*2v -2v[2x$e^y$+y$e^y$]

∂z/∂v = 2v[u²−v²]$e^{u^2-v^2}$-2v[2(u²+v²)$e^{u^2-v^2}$+y$e^{u^2-v^2}$]