Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

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.

share|improve this answer
    
@ 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 :

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

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

By the chain rule we know that :

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.