Answer
$\frac{\partial w}{\partial s}=4s,$ $\frac{\partial w}{\partial t}=4t$
For given values $s=1$ and $t=0$ is:
$\frac{\partial w}{\partial s}=4$ and $\frac{\partial w}{\partial t}=0.$
Work Step by Step
We will find both partial derivatives using the Chain Rule.
The partial derivative with respect to $s$ is:
$$\frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}=\frac{\partial}{\partial x}(x^2+y^2)\frac{\partial}{\partial s}(s+t)+\frac{\partial }{\partial y}(x^2+y^2)\frac{\partial}{\partial s}(s-t)=
2x\cdot1+2y\cdot1=2(x+y)$$
Expressing this in terms of $s$ and $t$ we get:
$$\frac{\partial w}{\partial s}=2(x+y)=2(s+t+s-t)=4s$$
The partial derivative with respect to $t$ is:
$$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}=\frac{\partial}{\partial x}(x^2+y^2)\frac{\partial}{\partial t}(s+t)+\frac{\partial}{\partial y}(x^2+y^2)\frac{\partial}{\partial t}(s-t)=2x\cdot1+2y\cdot(-1)=2(x-y)$$
Expressing this in terms of $s$ and $t$ we get:
$$\frac{\partial w}{\partial t}=2(x-y)=2(s+t-s+t)=4t$$
Now we will evaluate these partial derivatives at the given values $s=1$ and $t=0$:
$$\frac{\partial w}{\partial s}=4s=4\cdot1=4$$
$$\frac{\partial w}{\partial t}=4t=4\cdot0=0$$
