Answer
Verified
450.3k+ views
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.
\[d\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\,\, \times df\left( x \right) - f\left( x \right) \times dg\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\]
\[d\left( {f{\text{ }}\left( x \right).g{\text{ }}\left( x \right)} \right){\text{ }} = {\text{ }}f{\text{ }}\left( x \right).d{\text{ }}\left( {g{\text{ }}\left( x \right)} \right){\text{ }} + {\text{ }}d{\text{ }}\left( {f{\text{ }}\left( x \right)} \right).g{\text{ }}\left( x \right)\]
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.
\[d\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\,\, \times df\left( x \right) - f\left( x \right) \times dg\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\]
\[d\left( {f{\text{ }}\left( x \right).g{\text{ }}\left( x \right)} \right){\text{ }} = {\text{ }}f{\text{ }}\left( x \right).d{\text{ }}\left( {g{\text{ }}\left( x \right)} \right){\text{ }} + {\text{ }}d{\text{ }}\left( {f{\text{ }}\left( x \right)} \right).g{\text{ }}\left( x \right)\]
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE