Show that $\frac{\partial u}{\partial t} = a\frac{\partial u}{\partial x} + b\frac{\partial u}{\partial y} = 1$ 
If $u=f(r,s)$, $r=x+at$, $s=y+bt$ and $x$, $y$ and $t$ are independent variables and $a$ and $b$ are constants. Show that $\dfrac{\partial u}{\partial t} = a\dfrac{\partial u}{\partial x} + b\dfrac{\partial u}{\partial y} = 1$.

This is the question I was solving.
I know about partial derivative, but I'm unable to solve it.
I am unable to start the problem, I'm telling you my thoughts about the problem.
If $u$ is a function of $r$ and $s$, it's partial derivative w.r.t. $t$, $x$ and $y$ will be $0$ and hence $\dfrac{\partial u}{\partial t} = a\dfrac{\partial u}{\partial x} + b\dfrac{\partial u}{\partial y} = 0$ which is not equal to "$1$".
Can anyone help me?
 A: Using a tree diagram, we have with $u=f(r,s)$ and $\begin{cases}r=x+at,\\s=y+bt\end{cases}$ that,

*

*$\displaystyle  \frac{\partial u}{\partial t}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial u}{\partial s}\frac{\partial s}{\partial t}=\frac{\partial u}{\partial r}\cdot(a)+\frac{\partial u}{\partial s}\cdot (b)=a\frac{\partial u}{\partial r}+b\frac{\partial u}{\partial s}$.

*$\displaystyle \frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}=\frac{\partial u}{\partial r}\cdot (1)=\frac{\partial u}{\partial r}$.

*$\displaystyle \frac{\partial u}{\partial y}=\frac{\partial u}{\partial s}\frac{\partial s}{\partial y}=\frac{\partial u}{\partial s}\cdot (1)=\frac{\partial u}{\partial s}$.

Hence,
$$\color{blue}{\frac{\partial u}{\partial t}}=\color{red}{a\frac{\partial u}{\partial x}+b\frac{\partial u}{\partial y}}\iff \color{blue}{a\frac{\partial u}{\partial r}+b\frac{\partial u}{\partial s}}= \color{red}{ a\frac{\partial u}{\partial r}+b\frac{\partial u}{\partial s} }$$
So done
A: So you deal with a function $u$, with
$$u(x,y,t)=f(x+at,y+bt).$$
Therefore,
\begin{align*}\frac{\partial u}{\partial t}(x,y,t)&=(\partial_1 f)(x+at,y+bt)\frac{\partial }{\partial t}(x+at)+(\partial_2 f)(x+at,y+bt)\frac{\partial }{\partial t}(y+bt)\\
&=(\partial_1 f)(x+at,y+bt)a+(\partial_2 f)(x+at,y+bt)b.
\end{align*}
You have also
\begin{align*}\frac{\partial u}{\partial x}(x,y,t)&=(\partial_1 f)(x+at,y+bt)\frac{\partial }{\partial x}(x+at)+(\partial_2 f)(x+at,y+bt)\frac{\partial }{\partial x}(y+bt)\\
&=(\partial_1 f)(x+at,y+bt)
\end{align*}
and
\begin{align*}\frac{\partial u}{\partial y}(x,y,t)&=(\partial_1 f)(x+at,y+bt)\frac{\partial }{\partial y}(x+at)+(\partial_2 f)(x+at,y+bt)\frac{\partial }{\partial y}(y+bt)\\
&=(\partial_2 f)(x+at,y+bt).
\end{align*}
Therefore, you have your formula.
The fact that this is equal to $1$ should result from assumptions on $f$.
