$g=\sin$, $g'=\cos$
If $g(x)=\sin(x)$, then $g'(x)=\cos(x)$.
Then $g(y)=\sin(y)$ and $g'(y)=\cos(y)$.
Let $y=2x$.
Then $g(y)=\sin(2x)$ and $g'(y)=\cos(2x)$, but if $h(x)=\sin(2x)$, then $h'(x)=2\cos(2x)$,
so $h(x)=g(y)$, but $h'(x)$ is not equal to $g'(y)$ if, say, $x = \pi$.
This doesn't make sense to me, isn't to say $h(x)=g(y)$ to say that $h(x)$ and $g(y)$ are the same? and so wouldn't it follow that $h'(x)=g'(y)$?
Is my problem that I am looking at individual values, $h(x)$ and $g(y)$, instead of the functions $h$ and $g$ themselves?
Edit: Once I thought to think of derivative as slope and realized that the slope of the line tangent to the graph of $h(x)$ at $x=a$ is not necessarily the same as the slope of the line tangent to the graph of $g(x)$ at $x=a$, I understood (satisfactorily) why $g'(x)$ isn't the same as $h'(x)$.