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We have to find the derivative of $$f(x) = \dfrac{\tan(2x)}{\sin(x)}$$

I would like to know why my approach is incorrect:

$$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)} - \tan(2x) \cdot \cos(x)}{\sin^2(x)}$$

$$ = \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin^2(x)}$$

$$ = \dfrac {2 \sin(x) - \sin(2x) \cdot \cos(x)}{\cos^3(2x) \cdot \sin^2(x)}$$

p.s. - To avoid confusion ; I wanted to get rid of the $\tan$. I'm sure there is a shorter method than this but I don't want it; I just want to know why this is wrong.

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You've incorrectly simplified across the second equals sign, as well as the third. Be careful, $\frac{ab^{-1}-c}{d}\ne\frac{a-c}{bd}$. Can you see why? – Jared May 15 '13 at 21:05
If you are seeking to remove a denominator in the numerator or denominator of a compound fraction, you need to multiply across the entire numerator and denominator of that fraction by the factor you want to clear. Here, "top and bottom" need to be multiplied by $ \ \cos^2 (2x) \ $ in going to line 2 , leading to the result Amire Bendjeddou shows. ("Everybody" commits this oversight at one time or another...) – RecklessReckoner May 15 '13 at 21:25
@Jared , RecklessReckoner thanks, that was the mistake I was 'looking for'. – Derti May 15 '13 at 21:28
up vote 2 down vote accepted

Third line is: $ \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)\cdot\cos^2(2x)}{\cos^2(2x) \cdot \sin^2(x)}$ instead of $\\\\ \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin^2(x)}$

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$$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)} - \tan(2x) \cdot \cos(x)}{\sin^2(x)}\tag{1}$$

So far so good...

But your following line is where you made an algebraic error: you forgot to multiply both (the entire numerator (both terms)) and (the denominator) by $\cos^2(2x)$. Doing this gives us:

$$f'(x) = \dfrac{\color{blue}{\bf \cos^2(2x)}\left(\sin(x) \cdot \dfrac{2}{\cos^2(2x)} - \tan(2x) \cdot \cos(x)\right)}{\color{blue}{\bf \cos^2(2x)}\sin^2(x)}$$

NOW we distribute across the numerator, canceling the denominator of the first term, and multiplying the second term by $\cos^2{2x}$:

$$ = \dfrac{2\sin(x) - \tan(2x) \cdot \cos(x)\cdot \cos^2(2x)}{\cos^2(2x)\sin^2(x)}\tag{2}$$ $$ = \dfrac{2\sin(x) - \sin(2x) \cdot \cos^2(2x)\cdot \cos(x)}{\cos^3(2x)\sin^2(x)}\tag{3}$$

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Very nice use of color to make points +1 Today is a VERY slow day, but lots to do at work and prep for talk this Friday at work and talk next week. Hope you're having a great day (did you know they are trying to eliminate contractions)! – Amzoti May 16 '13 at 0:53

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