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This is a noob calculus question.

(1) $\sqrt{|xy|} = \sqrt[4]{x^2y^2}$, are the 2 expressions equal?

if yes to (1),

then why $\frac{d}{dx}\sqrt{|xy|} \neq \frac{d}{dx}\sqrt[4]{x^2y^2}$ ?

as $\frac{d}{dx}\sqrt{|xy|} =\frac{\sqrt[4]{x^2y^2}}{2x} $

but $\frac{d}{dx}\sqrt[4]{x^2y^2} = \frac{0.5xy^2}{(x^2y^2)^{0.75}}$

i obtained the above derivatives from wolfram alpha.

-updated- yup, i think they are just 2 different way of presentation, seems like they have the same graph, thanks for the clarification.

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I'm not clear on what your question is. Could you write it out a bit more? Are you asking why two things are equal and yet the third thing is not equal to them? Or are you asking if the first two things are equal? Or are you asking if the second two things are not equal? –  alex.jordan Nov 10 '11 at 3:24
    
@alex is this better? –  adsisco Nov 10 '11 at 3:28
2  
What suggests to you that $\frac{d}{dx}\sqrt{|xy|}\neq\frac{d}{dx}\sqrt[4]{x^2y^2}$? They are equal as far as I can tell. –  alex.jordan Nov 10 '11 at 3:53
1  
Also, "are the 2 equations equal" is a little nonsensical. Maybe you mean "are the 2 expressions equal"? –  alex.jordan Nov 10 '11 at 3:55

2 Answers 2

up vote 2 down vote accepted

They are indeed equal. This follows since

$$\begin{align} \sqrt[n]{x^n} = \left\{ \begin{array}{ccc} |x| & & n \text{ is even} \\ x & & n \text{ is odd} \end{array} \right. \end{align}$$

The derivatives are indeed equal (although the derivatives of neither function exist at $0$).

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For (1)

Since $ \sqrt{|xy|} = \sqrt[4]{|xy|^2} $ and $|xy|=\pm xy $

$\sqrt{|xy|} = \sqrt[4]{(\pm xy)^2} = \sqrt[4]{x^2y^2} $

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