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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?

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I don't think it is possible; d(AA)/dt= AdA/dt+ AdA/dt=2AdA/dt. – gary Nov 15 '11 at 6:08
Well I think so too... – wright Nov 15 '11 at 6:20

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.

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@ Magidin, What do you mean by Math Processing Error? \ – wright Nov 15 '11 at 6:21
@wright: It was an error in the LaTeX; it should have been corrected now. – Arturo Magidin Nov 15 '11 at 6:23

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 :

$\frac { d y}{ d u}=(u^2)'_u=2u=2f(x)$


By the chain rule we know that :

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}=2f(x)f'(x)$

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