Differentiate ${sin}^{2} 3 x . {tan}^{3} 2 x$

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Solution

$\frac{d}{d x} (s i n^{2} 3 x . t a n^{3} 2 x) = s i n^{2} 3 x \times \frac{d}{d x} (t a n^{3} 2 x) + t a n^{3} 2 x \times \frac{d}{d x} (s i n^{2} 3 x)$

$= s i n^{2} 3 x (3 t a n^{2} 2 x {sec}^{2} 2 x .2) + t a n^{3} 2 x (2 sin 3 x . c o s 3 x .3)$

$= 6 {sin}^{2} 3 x {tan}^{2} 2 x {sec}^{2} 2 x + 6 sin 3 x cos 3 x {tan}^{3} 2 x$

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