Question

Use the Chain Rule to find the indicated partial derivatives. z = x^4 + xy^3, x = uv^4 + w^3, y = u + ve^w Find : ∂z/∂u , ∂z/∂v , ∂z/∂w when u = 1, v = 1, w = 0

Answer 1

I'll use subscript notation for brevity, i.e.
`(\partial f)/(\partial x)=f_x`.

By the chain rule,

`z_u=z_xx_u+z_yy_u`

`z_v=z_xx_v+z_yy_v`

`z_w=z_xx_w+z_yy_w`

We have

`z=x^4+xy^3\implies\begin{cases}z_x=4x^3+y^3\\z_y=3xy^2\end{cases}`

and

`\begin{cases}x=uv^4+w^3\\y=u+ve^w\end{cases}\implies\begin{cases}x_u=v^4\\x_v=4uv^3\\x_w=3w^2\\y_u=1\\y_v=e^w\\y_w=ve^w\end{cases}`

When
`u=1,v=1,w=0`, we have

`\begin{cases}x(1,1,0)=1\\y(1,1,0)=2\end{cases}\implies\begin{cases}z_x(1,2)=12\\z_y(1,2)=12\end{cases}`

and the partial derivatives take on values of

`\begin{cases}x_u(1,1,0)=1\\x_v(1,1,0)=4\\x_w(1,1,0)=0\\y_u(1,1,0)=1\\y_v(1,1,0)=1\\y_w(1,1,0)=1\end{cases}`

So we end up with

`\boxed{\begin{cases}z_u(1,1,0)=24\\z_v(1,1,0)=60\\z_w(1,1,0)=12\end{cases}}`