finding a differentiable function $f(x,y)$ is a differentiable function satisfying the following properties:


*

*$f(x+t, y)= f(x,y) + ty$ and $f(x, y+t)= f(x,y) + tx$, $\forall x, y, t \in\mathbb{R}$ and

*$f(z, 0) = k$ for any $z\in\mathbb{R}$ and $k$ is an arbitrary constant. Find $f(x,y).$
 A: Use first principles:
$\frac{\partial f}{\partial x}=\lim_{t\to 0}\frac{f(x+t,y)-f(x,y)}{t}=\lim_{t\to 0}\frac{ty}{t}=y$ 
$\frac{\partial f}{\partial y}=\lim_{t\to 0}\frac{f(x,y+t)-f(x,y)}{t}=\lim_{t\to 0}\frac{tx}{t}=x$ 
Thus $f(x,y)=\int x dy=xy+g(x)$
and  $f(x,y)=\int y dx=xy+h(y)$
$f(z,0)=g(z)=h(0)=k$
So $g(x)=k, h(0)=k$, but also we need $h(y+t)=h(y)$, $\forall y,t\in\mathbb{R}\Rightarrow h(y)=k$
Thus $f(x,y)=xy+k$
A: Differentiate the two given equations with respect to $t$ to get
\begin{align}
f_1(x + t, y) & = y \\
f_2(x, y + t) & = x
\end{align}
where $f_i$ is the partial derivative with respect to the $i$-th parameter.
Plug in $t = 0$ to get
\begin{align}
f_1(x, y) & = y  \tag{1}\\
f_2(x, y) & = x. \tag{2}
\end{align}
Integrate (1) with respect to the first parameter to get
$$
f(x, y) = xy + c(y) \tag{3}
$$
where $c$ is some function of one variable.
Differentiate (3) with respect to $y$ to get $f_2(x, y) = x + c'(y)$. Equate with (2) to get $c'(y) = 0$. Therefore, $c(y)$ is a constant. Let's call it $c$. Now we get $f(x, y) = xy + c$. The condition $f(x, 0) = k$ makes $c = k$. Therefore, $f(x, y) = xy + k$ is a potential solution. Go back and check if this satisfies the two equations we started with.
