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

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Solution

Let $y = s i n^{2} 3 x t a n^{3} 2 x$
$\frac{d y}{d x} = 2 s i n 3 x . c o s 3 x . 3 t a n^{3} 2 x + s i n^{2} 3 x . 3 t a n^{2} 2 x . s e c^{2} 2 x .2$
$\Rightarrow \frac{d y}{d x} = 3 s i n 6 x t a n^{3} 2 x + 6 s i n^{3} 3 x t a n^{2} 2 x s e c^{2} 2 x$

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