If f: R rightarrow R is an even function havi | vedclass

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(B) Given that $f(x)$ is an even function, so $f(-x) = f(x)$.By differentiating both sides with respect to $x$ using the chain rule:$\frac{d}{dx}[f(-x

Vedclass · Accepted Answer

$f^{\prime}$

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(B) Given that $f(x)$ is an even function, so $f(-x) = f(x)$.By differentiating both sides with respect to $x$ using the chain rule:$\frac{d}{dx}[f(-x)] = \frac{d}{dx}[f(x)]$$-f^{\prime}(-x) = f^{\prime}(x)$, which implies $f^{\prime}(-x) = -f^{\prime}(x)$.Thus, the first derivative $f^{\prime}(x)$ is an odd function.Now, differentiating again:$\frac{d}{dx}[f^{\prime}(-x)] = \frac{d}{dx}[-f^{\prime}(x)]$$-f^{\prime \prime}(-x) = -f^{\prime \prime}(x)$, which implies $f^{\prime \prime}(-x) = f^{\prime \prime}(x)$.Thus, the second derivative $f^{\prime \prime}(x)$ is an even function.Similarly, the third derivative $f^{\prime \prime \prime}(x)$ will be an odd function.Therefore, $f^{\prime}(x)$ is an odd function. Comparing with the options, option $B$ is correct.

If $f: R \rightarrow R$ is an even function having derivatives of all orders, then which of the following is an odd function?

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