Question

What is the derivative of ${{\sin }^{2}}\left( 2x+3 \right)$?

Answer 1

You can use the chain rule to solve the question.
For applying “chain rule” we can consider an inner function and outer function.
In this case $\sin \left( 2x+3 \right)$ is the inner function.
The chain rule states that the derivative of a composite function is given by:
$F'\left( x \right)=f'\left( g\left( x \right) \right)\left( g'\left( x \right) \right)$
The derivation of $\sin x$ is $\cos x$ .
The derivative of ${{x}^{n}}=n{{x}^{n-1}}$
For the derivative of $2x+3$ since 3 is a constant its derivative is zero and derivative of $2x$ is 2.

Complete step by step solution:
Given,
${{\sin }^{2}}\left( 2x+3 \right)$
To find the derivative of ${{\sin }^{2}}\left( 2x+3 \right)$
Consider the derivative of the given function we can write the function as:
$F\left( x \right)={{\sin }^{2}}\left( 2x+3 \right)$
Thus, differentiate the above equation w.r.t $x$:
$F'\left( x \right)=2{{\sin }^{2-1}}\left( 2x+3 \right)\dfrac{d\left( \sin \left( 2x+3 \right) \right)}{dx}\times \dfrac{d}{dx}\left( 2x+3 \right)$
The derivation of $\sin x$ is $\cos x$ .
The derivative of ${{x}^{n}}=n{{x}^{n-1}}$
Applying chain rule to the equation we can write the derivative as mentioned above by considering the derivation of each possible term that can be differentiated.
$\begin{aligned} & F'\left( x \right)=2\sin \left( 2x+3 \right)\cos \left( 2x+3 \right)\times 2 \\\ & F'\left( x \right)=4\sin \left( 2x+3 \right)\cos \left( 2x+3 \right) \\\ \end{aligned}$

Hence the derivative of the function ${{\sin }^{2}}\left( 2x+3 \right)$ is $4\sin \left( 2x+3 \right)\cos \left( 2x+3 \right)$

Note: Another method to solve this question is that we know the formula listed below:
${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$
Thus, we can write the given function by using the formula stated above,
${{\sin }^{2}}\left( 2x+3 \right)=\dfrac{1-\cos 2\left( 2x+3 \right)}{2}$
Now the derivation of $\cos x$ is $-\sin x$
Therefore, we can write the derivative of the given function as:
$\begin{aligned} & \dfrac{d}{dx}\left( \dfrac{1-\cos 2\left( 2x+3 \right)}{2} \right) \\\ & =0-\dfrac{1}{2}\times 4\left( -\sin 2\left( 2x+3 \right) \right) \\\ & =2\sin 2\left( 2x+3 \right) \\\ \end{aligned}$
Since we know the formula of $\sin 2x=2\sin x\cos x$
Thus, we can write the above equation by using the formula of $\sin 2x$ we have,
$\begin{aligned} & 2\sin 2\left( 2x+3 \right)=2\left( 2\sin \left( 2x+3 \right)\cos \left( 2x+3 \right) \right) \\\ & =4\sin \left( 2x+3 \right)\cos \left( 2x+3 \right) \\\ \end{aligned}$
Thus, we see that the answer obtained in the above method and that explained earlier is the same thus any method could be used whichever is considered easy.
Do not forget to consider the negative sign for the derivative of $\cos x$.