How do you find the derivative of \[{{\sin }^{2}}\left( 2x+3 \right)\]?

Hint: This type of problem is based on the concept of trigonometry and differentiation. We have to use chain rule, that is, \[\dfrac{d}{dx}\left[ f\left( g\left( h\left( x \right) \right) \right) \right]={f}'\left( g\left( h\left( x \right) \right) \right){g}'\left( h\left( x \right) \right){h}'\left( x \right)\], where \[{f}'\left( g\left( h\left( x \right) \right) \right),{g}'\left( h\left( x \right) \right)\] and \[{h}'\left( x \right)\] are the derivative of f(g(h(x))), g(h(x)) and h(x) respectively. Here, \[f\left( g\left( h\left( x \right) \right) \right)={{\sin }^{2}}\left( 2x+3 \right)\] and g(h(x))=sin(2x+3) and h(x)=2x+3. Then, find the derivative of f(g(h(x))), g(h(x)) and h(x) and substitute those values in the chain rule to get the final answer.

Complete step-by-step solution:
According to the question, we are asked to find the derivative of \[{{\sin }^{2}}\left( 2x+3 \right)\].
We have been given the equation is\[{{\sin }^{2}}\left( 2x+3 \right)\]. ---------(1)
We can solve this question with the help of chain rule, that is,
\[\dfrac{d}{dx}\left[ f\left( g\left( h\left( x \right) \right) \right) \right]={f}'\left( g\left( h\left( x \right) \right) \right){g}'\left( h\left( x \right) \right){h}'\left( x \right)\], where \[{f}'\left( g\left( h\left( x \right) \right) \right),{g}'\left( h\left( x \right) \right)\] and \[{h}'\left( x \right)\] are the derivative of f(g(h(x))), g(h(x)) and h(x) respectively.
Let us compare the given function with chain rule and find f(g(h(x))), g(h(x)) and h(x).
First consider \[f\left( g\left( h\left( x \right) \right) \right)={{\sin }^{2}}\left( 2x+3 \right)\].
On differentiating \[f\left( g\left( h\left( x \right) \right) \right)\], we get
\[{f}'\left( g\left( h\left( x \right) \right) \right)=\dfrac{d}{dx}\left( {{\sin }^{2}}\left( 2x+3 \right) \right)\]
Using the power rule of differentiation, that is, \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\], we get
\[{f}'\left( g\left( h\left( x \right) \right) \right)=2\sin \left( 2x+3 \right)\dfrac{d}{dx}\left( \sin \left( 2x+3 \right) \right)\]
Now we have to find \[\dfrac{d}{dx}\left( \sin \left( 2x+3 \right) \right)\], that is, \[{g}'\left( h\left( x \right) \right)\].
\[{g}'\left( h\left( x \right) \right)=\dfrac{d}{dx}\left( \sin \left( 2x+3 \right) \right)\]
We know that \[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]. Therefore, we get
\[{g}'\left( h\left( x \right) \right)=\cos \left( 2x+3 \right)\dfrac{d}{dx}\left( 2x+3 \right)\]
Now, we have to find \[\dfrac{d}{dx}\left( 2x+3 \right)\] which is \[{h}'\left( x \right)\].
Therefore, \[{h}'\left( x \right)=\dfrac{d}{dx}\left( 2x+3 \right)\]
We know that \[\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}\]. Using this addition property of differentiation, that is, \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\], we get
\[{h}'\left( x \right)=2+\dfrac{d}{dx}\left( 3 \right)\]
Since the differentiation of a constant is always zero, \[\dfrac{d}{dx}\left( 3 \right)=0\].
Therefore, \[{h}'\left( x \right)=2\].
Now, substitute these values in \[\dfrac{d}{dx}\left[ f\left( g\left( h\left( x \right) \right) \right) \right]={f}'\left( g\left( h\left( x \right) \right) \right){g}'\left( h\left( x \right) \right){h}'\left( x \right)\], we get
\[\dfrac{d}{dx}\left( {{\sin }^{2}}\left( 2x+3 \right) \right)=2\sin \left( 2x+3 \right)\times \cos \left( 2x+3 \right)\times 2\]
We know that \[\sin A+\cos A=\sin 2A\]. Using this trigonometric identity, we get
\[\dfrac{d}{dx}\left( {{\sin }^{2}}\left( 2x+3 \right) \right)=2\sin 2\left( 2x+3 \right)\]
Therefore, the derivative of \[{{\sin }^{2}}\left( 2x+3 \right)\] is \[2\sin 2\left( 2x+3 \right)\].

Note: We can also solve this question by not substituting as f(g(h(x))), g(h(x)) and h(x). We should know the derivative of trigonometric functions to solve this type of question. Also, we should know some trigonometric identities to simplify the given function and the final answer. We should not make calculation mistakes based on sign conventions, if any.