Is the notation $f(x)′$ the same as $f′(x)$?

Question

$$f(x)=f(−x)⇒(f(x))′=(f(−x))′⇒f′(x)=−f′(−x)⇒f′$$

My question arises from this procedure.

Answer 1

Yes, $(f(x))'$ means $f'(x)$.

The tricky thing to remember, though, is that $f'(-x) \neq (f(-x))'$ in general. The former, $f'(-x)$, means to take the derivative first, and plug in $-x$ second. For example, $$f(x) = x^2 - x \implies f'(x) = 2x - 1 \implies f'(-x) = 2(-x) - 1 = -2x - 1.$$ The latter means plug in $-x$ first, then take the derivative, i.e. $$f(x) = x^2 - x \implies f(-x) = (-x)^2 - (-x) = x^2 + x \implies (f(-x))' = 2x + 1.$$ The proof uses the notation $(f(x))'$ to make it clear that the derivative with respect to $x$ is being taken on both sides of the equation.

Answer 2

Put $h(x):=f(-x)$. If $f(x)=h(x)$, then $f'(x)=h'(x)=-f'(-x)$

The notation $(f(x))'$ means $f'(x)$, but this is not a good notation.