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If $f(x)$ is an even function and $f^{\prime}(x)$ exists, then $f^{\prime}(e)+f^{\prime}(-e)$ is
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We have, $f(x)$ is an even function.
$$
\begin{array}{ll}
\therefore & f(-x)=f(x) \\
\Rightarrow & f^{\prime}(-x)(-1)=f^{\prime}(x) \\
\Rightarrow & f^{\prime}(x)+f^{\prime}(-x)=0 \\
\Rightarrow & f^{\prime}(e)+f^{\prime}(-e)=0
\end{array}
$$
$$
\begin{array}{ll}
\therefore & f(-x)=f(x) \\
\Rightarrow & f^{\prime}(-x)(-1)=f^{\prime}(x) \\
\Rightarrow & f^{\prime}(x)+f^{\prime}(-x)=0 \\
\Rightarrow & f^{\prime}(e)+f^{\prime}(-e)=0
\end{array}
$$
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