Differentiate \[{\sin ^2}3x.{\tan ^3}2x\,\]

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Hint:Use product rule $(f(x).g(x)$ to differentiate the value while differentiating ${\sin ^2}3x$ then first remove power and then solve the value. Use product rule \[\left( {f\left( x \right).g\left( x \right)} \right)\,\]

Complete step by step solution:
Let,\[y = {\sin ^2}3x.{\tan ^3}2x\,\]
Now, differentiate with respect to x by using product rule
\[\dfrac{{dy}}{{dx}} = \left( {\dfrac{d}{{dx}}\left( {{{\sin }^2}3x} \right)} \right)\,\,.\,\,\left( {{{\tan }^3}2x} \right) + {\sin ^2}3x\left( {\dfrac{d}{{dx}}\left( {{{\tan }^3}2x} \right)} \right)\]
\[ \Rightarrow \,\,2\,\,\sin \,\,3x\,\,.\,\,\cos \,3x\,\,.\,\,3\left( 1 \right)\,\,.\,\,\left( {{{\tan }^3}2x} \right)\]
\[ \Rightarrow \,\,2\,\,\sin \,\,3x\,\,.\,\,\cos \,3x\,\,.\,\,3\left( 1 \right)\,\,.\,\,\left( {{{\tan }^3}2x} \right) + \left( {{{\sin }^2}3x} \right)\,\,.\,\,3{\tan ^2}2x\,\,.\,\,{\sec ^2}2x\,\,.\,\,2\left( 1 \right)\]
$ \dfrac{{dy}}{{dx}} = 2\sin 3x.\dfrac{d}{{dx}}(\sin 3x).{\tan ^3}2x $ $ + {\sin ^2}3.3\tan 2x.\dfrac{d}{{dx}}\tan 2x. $
\[\dfrac{{dy}}{{dx}} = 2\sin 3x.\cos 3x.\dfrac{d}{{dx}}(3x) + {\tan ^3}2x + {\sin ^2}3x.\] $ 3.{\tan ^2}x.{\sec ^2}2x\dfrac{d}{{dx}}(2x) $
$ \dfrac{{dy}}{{dx}} = 2\sin 3x.\cos 3 \times 1.{\tan ^3}2x + {\sin ^2}3x.3{\tan ^2}2x{\sec ^2}2x.2(1) $
$ = \sin 3x.\cos 3x - {\tan ^3}2x + 6{\sin^2}3x.{\tan ^2}x{\sec ^2}2x $.

\[ \Rightarrow \,\,\,6\sin 3x\,\,\cos 3x\,\,{\tan ^3}2x + 6{\sin ^2}3x\,\,\,{\tan ^2}2x\,\,{\sec ^2}2x\]

Note: To solve this type of questions we always use a multiplication and division formula of differentiation.
\[d\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\,\, \times df\left( x \right) - f\left( x \right) \times dg\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\]
\[d\left( {f{\text{ }}\left( x \right).g{\text{ }}\left( x \right)} \right){\text{ }} = {\text{ }}f{\text{ }}\left( x \right).d{\text{ }}\left( {g{\text{ }}\left( x \right)} \right){\text{ }} + {\text{ }}d{\text{ }}\left( {f{\text{ }}\left( x \right)} \right).g{\text{ }}\left( x \right)\]