If $g$ is defined as $g(s)=f(t,y+s(x-y))$ why is $\displaystyle g'(s)=\frac{\partial f(t,y+s(x-y))}{\partial x}(x-y)$
Does it not have to be $\displaystyle g'(s)=\frac{\partial f(t,y+s(x-y))}{\partial (y+s(x-y))}(x-y)$
What, if it was $f(s+t,y+s(x-y))$ instead of $f(t,y+s(x-y))$ ?
It is at the top right
http://oi62.tinypic.com/110fz7l.jpg
The book: Ordinary Differential Equations by Luis Barreira and Claudia Valls
The function $f(t,x)$ depends on two variables: the first $t$ and the second $x$. The notation $\frac{\partial f}{\partial x}$ means the derivative with respect the second variable. When finding $$ \frac{d}{ds}f(t,y+s(x-y)) $$ we have by the chain rule $$\begin{align} \frac{d}{ds}f(t,y+s(x-y))&=\frac{\partial f}{\partial \text{ first variable}}\frac{dt}{ds}+\frac{\partial f}{\partial \text{ second variable}}\frac{d}{ds}(y+s(x-y))\\ &=\frac{\partial f}{\partial x}(t,y+s(x-y))\,(x-y). \end{align}$$