Question

If $f$ is an even function and $f^{\prime }\left(x\right)$ exists, then $f^{\prime }\left(0\right)=$

• A. $0$
• B. $1$
• C. $-1$
• D. $f\left(0\right)$

Answer 1

Answer: (A)

$0$

$f$ is an even function
$\therefore f\left(x\right)=f(-x)$
Differenting both sides, we get
$f^{\prime }\left(x\right)=-f^{\prime }(-x)$
It is given that $f^{\prime }\left(x\right)$ exist, implies that $f^{\prime }\left(0\right)$ also exists.
$\Rightarrow f^{\prime }\left(0\right)=-f^{\prime }\left(0\right)$
$\Rightarrow 2f^{\prime }\left(0\right)=0$
$\Rightarrow f^{\prime }\left(0\right)=0$