I have seen it in an exercise book. I don't know how to do it.
If $f(x)$ is continuous at $x=a,$ and $|f(x)|$ can be differentiated at $x=a,$ then $f(x)$ is differentiable at $x=a.$
I have seen it in an exercise book. I don't know how to do it.
If $f(x)$ is continuous at $x=a,$ and $|f(x)|$ can be differentiated at $x=a,$ then $f(x)$ is differentiable at $x=a.$
Let $g(x)=|f(x)|$. If $f(a)\ne 0$, then by the continuity of $f$, there is an interval $I_\epsilon=(a-\epsilon, a+\epsilon)$ such that if $x$ is in $I_\epsilon$, then $f(x)$ has the same sign as $f(a)$. If $f(a)$ is positive, then for all $x$ in $I_\epsilon$, we have $g(x)=f(x)$. So for $x$ in the interval $I_\epsilon$, $\frac{g(x)-g(a)}{x-a}=\frac{f(x)-f(a)}{x-a}.$ It follows that if $\lim_{x\to a}\frac{g(x)-g(a)}{x-a}$ exists, then so does $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ (and the limits are the same). We conclude that $f'(a)$ exists and is equal to $g'(a)$.
If $f(a)$ is negative, then for all $x$ in $I_\epsilon$, we have $g(x)=-f(x)$. It follows that $\frac{g(x)-g(a)}{x-a}=\frac{-f(x)+f(a)}{x-a}=-\frac{f(x)-f(a)}{x-a}.$ Thus if $\lim_{x\to a}\frac{g(x)-g(a)}{x-a}$ exists, then so does $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$. In this case, $f'(a)=-g'(a)$.
Now we come to the interesting case, where $f(a)=0$. We are told that $\lim_{x\to 0}\frac{g(x)-g(a)}{x-a}$ exists. Since $g(a)=0$, this says that $\lim_{x\to a}\frac{g(x)}{x-a}$ exists.
But $g(x)\ge 0$ for all $x$. So for all $x>a$, we have $\frac{g(x)}{x-a}\ge 0$. It follows that $\lim_{x\to a+}\frac{g(x)}{x-a} \ge 0.\tag{$1$}$ Also, if x, then x-a<0, and therefore $\frac{g(x)}{x-a}\le 0$. It follows that $\lim_{x\to a-}\frac{g(x)}{x-a} \le 0.\tag{$2$}$ Since by assumption $\lim_{x\to a}\frac{g(x)}{x-a}$ exists, we conclude from inequalities $(1)$ and $(2)$ that $\lim_{x\to a}\frac{g(x)}{x-a} = 0.\tag{$3$}$ It follows immediately that $\lim_{x\to a}\frac{f(x)}{x-a} = 0.$ This says that $f'(a)$ exists and is equal to $0$.