Why is the differentiability of a piecewise defined function studied using the definition of derivative? Why is the differentiability of a piecewise defined function often studied using the definition of derivative?
For example, let:
$$\begin{align*}
 f(x)
&= x\cdot |x-1|\\
&= \left\{ \begin{array}{lcl}x (-x+1)& \text{if} & x <
1 \\ x  (x-1) & \text{if} & x \geq 1 \end{array} \right.\\
&= \left\{ \begin{array}{lcl}-x^2+x& \text{if} & x <
1 \\ x^2-x & \text{if} & x \geq 1 \end{array} \right.
\end{align*}$$
$f$ is continuous in $\mathbb{R} $ except, maybe, when $x=1$:
$$\begin{align*}
 \lim_{x \to 1^{-}} f(x) &= \lim_{x \to 1} (-x^2+x) = 0\\
\lim_{x \to
1^{+}} f(x)&= \lim_{x \to 1} (x^2 -x) = 0 \end{align*}$$
$f$ is continuous when $x=1$ because:
$$ \lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{+}} f(x) = f(1) $$
As to the differentiability at that point (by derivative's definition):
$$\begin{align*}
\lim_{h \to 0^{-}} \frac{f(1+h)-f(1)}{h} &= \lim_{h \to 0}
\frac{\left[-(1+h)^2 + (1+h)\right] - \left[-1^2 + 1\right]}{h}\\
 &= \lim_{h\to 0}
\frac{- h(h+1)}{h} = -1\\
\strut\\
 \lim_{h \to 0^{+}} \frac{f(1+h)-f(1)}{h} &= \lim_{h \to 0} \frac{\left[(1+h)^2
- (1+h)\right] - \left[1^2 - 1\right]}{h}\\ 
&= \lim_{h\to 0} \frac{h(h+1)}{h} = 1 
\end{align*}$$
$$ \lim_{h \to 0^{-}} \frac{f(1+h)-f(1)}{h} \neq \lim_{h \to 0^{+}} \frac{f(1+h)-f(1)}{h} $$
so $f$ is not differentiable when $x=0$.
The "second" method is:
$$ f'(x) = \left\{ \begin{array}{lcl} -2x + 1& \text{if} & x <
1 \\ 2x-1 & \text{if} & x > 1 \end{array} \right. $$
$$ f'(1^{-}) = -1 \neq f'(1^{+}) = 1 $$
Is there any case in which the differentiability study by using the derivative (the second method) would be wrong?

Consider the function
$$f(x) = \left\{\begin{array}{ll}
x^2\sin\left(\frac{1}{x}\right) &\text{if }x\neq 0,\\
0 &\text{if }x=0.
\end{array}\right.$$
The function is differentiable at $0$, with derivative equal to $0$:
$$\begin{align*}
\lim_{h\to 0}\frac{f(h)-f(0)}{h} &= \lim_{h\to 0}\frac{h^2\sin(\frac{1}{h})}{h} \\
&=\lim_{h\to 0}h\sin(\frac{1}{h}) = 0.
\end{align*}$$
However, using the "second method", we look at the derivative of $x^2\sin(\frac{1}{x})$:
$$\left(x^2\sin\left(\frac{1}{x}\right)\right)' = 2x\sin\frac{1}{x} - \cos\frac{1}{x}.$$
And
$$\lim_{x\to 0}\left(2x\sin\frac{1}{x}-\cos\frac{1}{x}\right)$$
does not exist.
Using the method, we would incorrectly conclude that the derivative does not exist at $0$.
In general, a derivative does not have to be continuous; so the second method may yield "false negatives", functions that are differentiable but which the technique suggests are not.
