Question

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

Answer 1

In this question we have to find the derivative of the given equation.
We can solve the given function to find the derivative of the above given function by using differentiation rules. First let us denote the given function as $f\left( x \right)$ and then rewrite the given function to derive it easier.
And then differentiate the given function by using power and other differentiation rules to get the result. You will also have to use chain rule to differentiate further.
Hence we can get the required derivative.

Complete step-by-step solution:
Let us take
$f\left( x \right) = {\sin ^2}\left( {2x + 3} \right)\,$ …. (1)
The above equation can be written in the form as follows
$f\left( x \right) = {\left( {\sin \left( {2x + 3} \right)} \right)^2}$ …. (2)
Now differentiating the above function we can get the result as follows,
$f'\left( x \right) = 2\sin \left( {2x + 3} \right)\dfrac{d}{{dx}}\left[ {\sin \left( {2x + 3} \right)} \right]\,\,\,\,\,\,\left[ {\because \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}} \right]$ …. (3)
We can differentiate $\left[ {\sin \left( {2x + 3} \right)} \right]$ directly in the above step, but we are going to differentiate it separately to make it easier and to avoid mistakes.
Now Consider, $\dfrac{d}{{dx}}\left[ {\sin \left( {2x + 3} \right)} \right]$
Again differentiate we get,
$\dfrac{d}{{dx}}\left[ {\sin \left( {2x + 3} \right)} \right] = \cos \left( {2x + 3} \right)\dfrac{d}{{dx}}\left[ {\left( {2x + 3} \right)} \right]\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x} \right]$
$\dfrac{d}{{dx}}\left[ {\sin \left( {2x + 3} \right)} \right] = \cos \left( {2x + 3} \right)\left( 2 \right)\,\,\,\,\,\,\,\,\left[ {\because \dfrac{d}{{dx}}\left( {2x} \right) = 2\left( 1 \right) = 2} \right]$
$\dfrac{d}{{dx}}\left[ {\sin \left( {2x + 3} \right)} \right] = 2\cos \left( {2x + 3} \right)$ …. (4)
Substitute equation $\left( 4 \right)$ in equation $\left( 3 \right),$ we get
$f'\left( x \right) = 2\sin \left( {2x + 3} \right)\left( {2\cos \left( {2x + 3} \right)} \right)$
Simplifying the equation,
$f'\left( x \right) = 4\sin \left( {2x + 3} \right)\left( {\cos \left( {2x + 3} \right)} \right)$
Using $\sin 2A = 2\sin A\cos A$
$f'\left( x \right) = 2\sin 2\left( {2x + 3} \right)\,$
$f'\left( x \right) = 2\sin \left( {4x + 6} \right)$
$\dfrac{d}{{dx}}{\sin ^2}\left( {2x + 3} \right) = 2\sin \left( {4x + 6} \right)$
Using power rule and other differentiation rules we can get the required derivative of a function.

Thus, the derivative of ${\sin ^2}\left( {2x + 3} \right)$ is $2\sin \left( {4x + 6} \right)$ .

Note: We can also solve this question by the following method
${\sin ^2}\left( {2x + 3} \right) = \sin \left( {2x + 3} \right)\sin \left( {2x + 3} \right)$
By the differentiation rule
$\dfrac{d}{{dx}}f\left( x \right)g\left( x \right) = f'\left( x \right)g\left( x \right) + f\left( x \right)g'\left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( * \right)$
And by using the chain rule,
We differentiate the function respect to them
$\dfrac{d}{{dx}}{\sin ^2}\left( {2x + 3} \right) = \dfrac{d}{{dx}}\sin \left( {2x + 3} \right)\sin \left( {2x + 3} \right)$
Here $f\left( x \right) = \sin (2x + 3),g\left( x \right) = \sin \left( {2x + 3} \right)$
Substitute in equation $\left( * \right)$
$\dfrac{d}{{dx}}\sin \left( {2x + 3} \right)\sin \left( {2x + 3} \right) = \left[ {\dfrac{d}{{dx}}\sin \left( {2x + 3} \right)} \right]\sin \left( {2x + 3} \right) + \sin \left( {2x + 3} \right)\left[ {\dfrac{d}{{dx}}\sin \left( {2x + 3} \right)} \right]$
Using $\dfrac{d}{{dx}}\sin x = \cos x\,,\,\dfrac{d}{{dx}}nx = n$ ,
$\Rightarrow \dfrac{d}{{dx}}\sin \left( {2x + 3} \right)\sin \left( {2x + 3} \right) = \cos \left( {2x + 3} \right)\left( 2 \right)\sin \left( {2x + 3} \right) + \sin \left( {2x + 3} \right)\cos \left( {2x + 3} \right)\left( 2 \right)\,\,\,\,\,\,\,\,\,$
$\Rightarrow \dfrac{d}{{dx}}\sin \left( {2x + 3} \right)\sin \left( {2x + 3} \right) = 4\cos \left( {2x + 3} \right)\sin \left( {2x + 3} \right)$
$\Rightarrow \dfrac{d}{{dx}}\sin \left( {2x + 3} \right)\sin \left( {2x + 3} \right) = 2\sin 2\left( {2x + 3} \right)\,\left[ {\because \sin 2A = 2\sin A\cos A} \right]$
$\Rightarrow \dfrac{d}{{dx}}\sin \left( {2x + 3} \right)\sin \left( {2x + 3} \right) = 2\sin \left( {4x + 6} \right)$
$\Rightarrow \dfrac{d}{{dx}}{\sin ^2}\left( {2x + 3} \right) = 2\sin \left( {4x + 6} \right)$
Thus the derivative of ${\sin ^2}\left( {2x + 3} \right)$ is $2\sin \left( {4x + 6} \right)$ .
The derivative of a function $y = f\left( x \right)$ of a variable $x$ is a measure of the rate at which the value of y of the function changes with respect to the change of the variable$x$ . It is called the derivative of $f$ with respect to$x$ .
There are various methods to solve the derivative of the function.
Some of them are as follows,
The chain rule, the product rule, the quotient rule, implicit differentiation, derivative of trigonometric functions, derivative of exponential functions, derivative of logarithms.
We can easily find the derivative of a function, when we have known all the formulas of the differentiation.