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.
Answer
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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