Derivative Notation: Let $f(x)=x$. Is $f′(x^2)=1$ or is $f′(x^2)=2x$?

Question

Let $f(x)=x$.

Is $f'(x^2)=1$ or is $f'(x^2)=2x$?

In other words, using this notation, are we evaluating the derivative at a $x^2$?

Answer 1

If you consider the function $f$ of a variable $x$, then $f'$ is likewise a function. When you write $f'(a)$, you evaluate that function at $a$. If $a$ happens to be $x^2$, then you evaluate at $x^2$. It has nothing to do with differentiating with respect to $a$ or $x^2$ or anything else. So, in your example, $f'(a)=1$ no matter what you put in for $a$ when $f(x)=x$.

If you consider $g(x)=x^2$ and the composition $(f\circ g)(x) = f(x^2)$, then $(f\circ g)'(x) = f'(g(x))g'(x) = f'(x^2)\cdot 2x$.

Answer 2

It seems like you're getting hung up on the notation, which is understandable because this notation can mean different things in different contexts.

$f(x)$ is always a function of the variable $x$.

$f'(x)$ is the derivative of $f(x)$. Perhaps you could think of this as saying $f$ has already been differentiated so all we're interested in is its argument. The outermost function of $f'(x^2)$ and $f'(\sqrt{x})$ is the same, i.e. $f'(\cdot)$; the only thing changing is whether we substitute $x^2$ or $\sqrt{x}$. For example, if $f(x)=x^3$, then $f'(x)=3x^2$ and $f'(x^2)=3(x^2)^2$. This explains how Ted's answer has the derivative of $F(x)=(f\circ g)(x)=x^2$ computed as $F'(x)=(f\circ g)'(x)=2x$. Note that the term $f'(x^2)$ was (unintuitively) $1$, since $f'(x)=1$ for all $x$.

Conversely, $\left(f(x)\right)'$ has the derivative on the outside so it is the last step. So this puts more emphasis on differentiating what's inside the brackets. For instance $\left(f(x^2)\right)'$ would usually mean that we've taken the function $f(x^2)$ and are then differentiating that wrt. $x$. This is different from $f'(x^2)$, where we first differentiated and then substituted $x^2$. For example, if $f(x)=x^3$, then $f(x^2)=(x^2)^3=x^6$ and $\left(f(x^2)\right)'=6x^5$. This explains how Yves' answer has the derivative of $F(x)=x^2$ computed as $F'(x)=(x^2)'=2x$.

Hopefully this addresses your original question. When we define $f(x)=x$, the expression $f'(x^2)$ means first take the derivative ($=1$) and then substitute $x^2$, which makes no difference in this case since the derivative is constant.