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$g=\sin$, $g'=\cos$

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 $h'(x)$.

  • 6
    You are using primes to denote two different things, which we can loosely describe as "derivatives with respect to x" and "derivatives with respect to y"2012-10-14
  • 0
    @Mariano: The primes mean the same thing in both cases: "The derivative of the univariate function that is decorated with the prime".2012-10-14
  • 0
    they cannot mean that, for otherwise, for example, the equalities $g(y)=\sin(2x)$ and $g'(y)=\cos(2x)$ are simply incompatible.2012-10-14
  • 0
    What's the problem? $g' = \cos$, so $g'(y) = \cos y = \cos(2x)$2012-10-14
  • 0
    Hurkyl is correct about the meaning of "prime". However, there is implicit equivocation involved in the OP's conclusion. (See my answer.)2012-10-14

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