Problem of the day

As $f$ is as $\color{#20A900}{even function}$ ,
$\therefore \large \color{#3D99F6}{f(x) = f(-x)}$
Differentiating both sides, we get
$f_1(x)$ = $-f_1(-x)$
It is given that $f_1(x)$ exists $\color{#EC7300}{\Rightarrow}$ $f_1(0)$ also exists.
$\Rightarrow$ $f_1(0)$ = $-f_1(0)$
$\Rightarrow$ 2 $f_1(0)$ = 0
$\Rightarrow \color{#69047E}{\boxed{f_1(0) = 0}}$

f(x) is an even function implies it contains only even powers of x and some constant .f'(x) means first derivative of f(x) . f'(x) will contain only odd powers of x and no constant term , hence f'(0) is 0