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f'(0) = ? for an Even Function f Differentiable at 0

Since f is differentiable at 0 we have,

$\mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 - h) - f(0)}}{{ - h}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h} =  - \mathop {\lim }\limits_{h \to 0} \frac{{f( - h) - f(0)}}{h}$

Since f is an even function, f(-h) = f(h)

$\therefore \mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h} =  - \mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h}$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h} = 0$

$ \Rightarrow 2f'(0) = 0$

$\therefore f'(0) = 0$