I've seen someone claim that $g(x) = f(x) - f(-x) => g'(-x) = f'(x) -(-1 * f'(x))$ by the chain rule. I just can't grasp my head around why is this true?
2026-04-05 20:15:56.1775420156
Question about the chain rule and derivates.
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No, that (was originally $f'(-x)=-1·f'(x)$) is not true, just take $f(x)=x$ where you have a constant derivative $f'(x)=1$.
What you can however do is define $f_-(x)=f(-x)$ where then $$f_-'(x)=-f'(-x)$$ and changing the point of the derivative evaluation $$f_-'(-x)=-f'(x).$$
In the modified question, the odd function $g(x)=f(x)-f(-x)$ has the even-symmetric derivative $$ g'(x)=f'(x)+f'(-x) $$ which remains unchanged under changing $x$ to $-x$ in all places.