Product rule in calculus This is wonderful question I came across whiles doing calculus. We all know that $$\frac{d(AB)}{dt} = B\frac{dA}{dt} + A\frac{dB}{dt}.$$ 
Now if $A=B$ give an example for which 
$$\frac{dA^2}
{dt} \neq 2A\frac{dA}{at}.$$ 
I have tried many examples and could't get an example, any help?
 A: If $A=|t|$, then $A^2 = t^2$; so $\frac{dA^2}{dt} = \frac{d}{dt}t^2 = 2t$ for all $t$.
On the other hand,
$$\begin{align*}
2A\frac{dA}{dt} &= \left\{\begin{array}{ll}
2|t|&\text{if }t\gt 0;\\
2|t|(-1)&\text{if }t\lt 0
\end{array}\right.\\
 &= 2t,\quad t\neq 0.\end{align*}$$
So they are equal where they are both defined, but not equal at $t=0$,as $2A\frac{dA}{dt}$ does not exist there.
For a more radical example, take
$$A(t) = \left\{\begin{array}{ll}1 &\text{if }t\in\mathbb{Q},\\
-1&\text{if }t\notin\mathbb{Q}.
\end{array}\right.$$
Then $A(t)$ is not continuous anywhere, so the derivative does not exist anywhere; however, $(A(t))^2 = 1$ for all $t$, so the derivative always exists (and is equal to $0$). Looking at
$$\frac{d}{dt}A^2 = 2A\frac{dA}{dt},$$
the left hand side makes sense, but the right hand side does not (since $\frac{dA}{dt}$ does not exist). 
The key point here is that the Product Rule assumes that both factors are differentiable. It is possible for a product to be differentiable and yet for each factor to not be differentiable. In that situation, the Product Rule does not apply.
A: let's observe function 
$y=(f(x))^2$ , this function can be decomposed as the composite of two functions:
$y=f(u)=u^2$ and $u=f(x)$
So :
$\frac { d y}{ d u}=(u^2)'_u=2u=2f(x)$
$\frac{du}{dx}=f'(x)$
By the chain rule we know that :
$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}=2f(x)f'(x)$
