If f : R to R is an even function, then which | vedclass

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(D) $1$. By definition, a function $f(x)$ is even if $f(-x) = f(x)$ for all $x \in R$.$2$. Differentiating both sides with respect to $x$ using the ch

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$f'(x)$ is an odd function

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(D) $1$. By definition, a function $f(x)$ is even if $f(-x) = f(x)$ for all $x \in R$.$2$. Differentiating both sides with respect to $x$ using the chain rule: $\frac{d}{dx}[f(-x)] = \frac{d}{dx}[f(x)]$.$3$. This gives $f'(-x) \cdot (-1) = f'(x)$, which simplifies to $f'(-x) = -f'(x)$.$4$. $A$ function $g(x)$ is odd if $g(-x) = -g(x)$. Since $f'(-x) = -f'(x)$, $f'(x)$ is an odd function.

If $f : R \to R$ is an even function, then which of the following is true?

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