I have seen in many examples that using the fundamental definition of derivative only gives the correct results while checking for differentiability at a point. But why do we get wrong results while we directly apply the rules of differentiation and taking the limit at the point. Here is an example
$$y = \begin{cases}
x^2, &-\infty < x < 1\\
2x, &1\leq x < \infty
\end{cases}
$$
If we check the differentiability at the point $x=1$ then when we apply the rules of differentiation left hand and right hand derivative becomes same. But by using fundamental definition of derivative it will be different.
But if I am right the rules of differentiation are in fact derived from the fundamental definition of derivative? So both of these should give correct result. Or it seems like I am missing some basic concept.
I have seen same question like this one here. The answer to this question only says it gives wrong result.But it doesn't answer the question of why?
Best Answer
The issue here is that the derivative $y'(1)$ is not defined. I'll start with the definition and then discuss why the differentiation rules don't apply at $x = 1$. First, for the derivative to exist at $x = 1$, then by definition,the following limit has to exist: $$\lim_{h\to 0}\frac{y(1+h) - y(1)}{h} \tag{1}.$$ Now, in order for $(1)$ to exist the left and right limits have to exist and agree. So, we can look at the left and right derivatives:
Right Derivative: $$y'_{+}(1) = \lim_{h\to 0^{+}}\frac{f(1+h) - f(1)}{h} = \lim_{h\to 0^{+}}\frac{2(1+h) - 2}{h} = \lim_{h\to 0^{+}}\frac{2h}{h} = 2$$ There's no trouble there, however we run into problems with the left derivative:
Left Derivative: \begin{align} y'_{-}(1)&=\lim_{h\to 0^{-}}\frac{y(1+h) - y(1)}{h}\\ &= \lim_{h\to 0^{-}}\frac{(1+h)^2 - 2}{h}\\ &= \lim_{h\to 0^{-}}\frac{1 +2h+h^2-2}{h}\\ &= \lim_{h\to 0^{-}}\frac{h^2+2h-1}{h}\\ &= \lim_{h\to 0^{-}}\left(h + 2 - \frac{1}{h}\right)\\ &= \infty \end{align}
So, since the left and right derivatives don't agree, we can say that $y$ is not differentiable at $y = 1$. As has been pointed out in the comments we could also have concluded that from continuity, because if $y'(1)$ had existed, then $y$ would need to be continuous at $x = 1$, and because it is not we can immediately rule out differentiability there.
Now, to understand why it doesn't work to take the derivatives of $2x$ and $x^2$ and take a limit, we start by noting that we can certainly write $$y'(x) = \begin{cases} 2x,&-\infty<x<1\\ 2, &1<x<\infty \end{cases}$$
However, if we were to write $$y'(1) = \lim_{x\to 1}y'(x)$$ we would be making two unfounded assumptions:
First, that $y$ is actually differentiable at $x = 1$, and so writing $y'(1)$ makes sense, and second that $y'$ is continuous. We don't know a priori that either of those things are true, and in fact as previously mentioned, we know that $y'(1)$ can't exist because $y$ isn't continuous at $x = 1$.
To give a bit more of an intuitive explanation, remember that the derivative at a point relies on the value of the function at that point. So, when we use $2x$ for the derivative on the left, instead of using $y(1) = 2$, which lies on the right piece, we're really using $\lim_{x\to 1^{-}}y(x) = 1$ on the left piece.