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In a problem Spivak mentions that $f'(x^2)$ means the derivative of $f$ at the number which we happen to be calling $x^2$; it is not the derivative at $x$ of the function $g(x) = f(x^2)$.

This sentence really struck me because it means that I have been doing all of the Spivak's previous problems in a wrong way.

For example, when doing the problem: prove that $\lim_{x\to 0^+} f(x) = \lim_{x\to 0} f(x^2)$, I used the following: $f(x^2) = (g \circ h) (x)$, where $g(x) = f(x)$ and $h(x) = x^2$.

Why am I so shocked: I always used this notation and I was always able to prove everything I needed - I have never found a contradiction.

So, how to think about $f(x^2)$ ? And if it really just means that it is the value of the function evaluated at some number $x^2$ then why write it like a square at all (I assume that probably to say that the numbers considered are positive) and how to interpret $\lim_{x\to 0} f(x^2)$?

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    $\begingroup$ You have not been doing the problems wrong, based on the example you showed. It's true and correct that $f(x^2) = (g \circ h)(x)$, where $g(x) = f(x)$ and $h(x) = x^2$. Why would Spivak's statement about $f'(x^2)$ imply that this isn't true? [And based on your previous correct understanding, why would you think that $f'(x^2)$ is the derivative at $x$ of the function $g(x) = f(x^2)$?] $\endgroup$ – littleO Apr 28 '17 at 9:35
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    $\begingroup$ But the function $f$ is defined as : $f(x) := x^3$. So its derivative is defined: $f'$ must be $3x^2$. $\endgroup$ – Mauro ALLEGRANZA Apr 28 '17 at 9:41
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    $\begingroup$ Thus, the "tricky" exercise ask to compute (evaluate) $f'$ (that is a known function) at different "points" : $9, 3^2,a^2,x^2$. $\endgroup$ – Mauro ALLEGRANZA Apr 28 '17 at 9:43
  • $\begingroup$ Thank you for your comments! Do I understand correctly that if $g(x) = f'(x)$ and $h(x) = x^2$ then $f'(x^2) = (g \circ h)(x)$ and that's it? $\endgroup$ – Daniels Krimans Apr 28 '17 at 9:59
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    $\begingroup$ Both you and Spivak are correct. The notation $f'(x)$ where $x$ is a variable means $\dfrac{d} {dx} f(x) $ and hence $f'(g(x)) $ means $(f'\circ g) (x)$. It does not mean $(f\circ g) '(x) $. $\endgroup$ – Paramanand Singh Apr 28 '17 at 10:26
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$f'(x^2)$ is indeed not the same as $(f \circ h)'(x)$. Its maybe easier to be confused if you use $\frac{d}{dx}$ notation, and write $f(x)$ with $x$ 'indeterminate' to mean the function $f = f(x)$, giving (in general) the non-identity $$ \frac{df}{dx}(x^2) \neq \frac{d}{dx} (f(x^2))$$

In the case of the limit $\lim_{x→ 0} f(x^2)$ there is no issue because there is no way to ambiguously bracket the expression.

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