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 $x$,we get $f^{\prime}(-x) \cdot (-1) = f^{\

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$f^{\prime}$

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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 $x$,we get $f^{\prime}(-x) \cdot (-1) = f^{\prime}(x)$,which implies $f^{\prime}(-x) = -f^{\prime}(x)$. Thus,$f^{\prime}$ is an odd function.By differentiating again,$f^{\prime \prime}(-x) \cdot (-1) = -f^{\prime \prime}(x)$,which implies $f^{\prime \prime}(-x) = f^{\prime \prime}(x)$. Thus,$f^{\prime \prime}$ is an even function.By differentiating once more,$f^{\prime \prime \prime}(-x) \cdot (-1) = f^{\prime \prime \prime}(x)$,which implies $f^{\prime \prime \prime}(-x) = -f^{\prime \prime \prime}(x)$. Thus,$f^{\prime \prime \prime}$ is an odd function.Among the given options,$f^{\prime}$ is an odd function. Therefore,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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