Can we take individual derivative of piecewise function if the function is continuous and differentiable?

Question

Can we take individual derivative of piecewise function if the function is continuous and differentiable?

Suppose a function $f(x)$ is defined in such a way that it's definition changes at some particular point, say $x = a$. If it's given a priori that $f(x)$ is continuous and differentiable at point $x = a$, can we just individually take derivative of one of the function and find it's value at $x = a$?

For example, $$f(x) = \begin{cases} x^2 & \text{if}\ x \geq 1 \\ 2x - 1 & \text{otherwise} \end{cases}$$

This function is continuous and differentiable at $x=1$. Here the derivative also equals $$2 = 2(1) \text{ (derivative of } x^2 \text{ at 1}) = 2 \text{ (derivative of } 2x \text{ at 1})$$

Is this true for all functions given to be continuous and differentiable at some point (that value of individual derivative equals actual derivative)?

Answer 1

Yes, that is (almost) true. For: $$ f(x)= \begin{cases} g(x) &\text{if }x\geq a\\ h(x) &\text{otherwise} \end{cases} $$ and both $g,h$ are defined and continuous at $x=a$, then we must have: $$ f\text{ is continuous at }a\iff g(a)=h(x) $$ Now regarding $f$ being differentiable at $a$, we have: $$ \text{$g$ and $h$ are differentiable at $a$}\\ \quad\\ \Downarrow\\ \quad\\ \begin{bmatrix} g'(a)=h'(a)=\alpha\\ \Updownarrow\\ f\text{ is differentiable at }a\\ \text{ and }\\ f'(a)=\alpha \end{bmatrix} $$

BUT if either $g$ or $h$ is not differentiable at $a$ (or for instance $h$ is not defined there), then you need to check the actual definition of being differentiable in detail to figure out what happens there.