Question

How do you use the chain rule of differentiation $y = {\sin ^3}\left( {2x + 1} \right)$?

Answer 1

Hint: This problem deals with differentiation and also trigonometry of the given equation using chain rule. The chain rule tells us how to find the derivative of a composite function. It is applied on composite functions. That is if a function is a product of two or more functions, then the differentiation of the composite function is given by the chain rule of differentiation. The chain rule is given below:
$ \Rightarrow \dfrac{d}{{dx}}\left( {{f_1}(x).{f_2}(x)} \right) = {f_1}(x).\dfrac{d}{{dx}}\left( {{f_2}(x)} \right) + {f_2}(x).\dfrac{d}{{dx}}\left( {{f_1}(x)} \right)$

Complete step by step answer:
Given an equation of trigonometric function, which is $y = {\sin ^3}\left( {2x + 1} \right)$
Let $y = f(x)$
If $f(x) = {f_1}(x) + {f_2}(x)$
We know that the differentiation of the above equation gives:
$ \Rightarrow \dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}{f_1}(x) + \dfrac{d}{{dx}}{f_2}(x)$
Now consider the given function, $f(x) = {\sin ^3}\left( {2x + 1} \right)$ as shown below:
$ \Rightarrow f(x) = {\sin ^3}\left( {2x + 1} \right)$
Now differentiate the above equation as shown below:
\[ \Rightarrow \dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}\left( {{{\sin }^3}\left( {2x + 1} \right)} \right)\]
Here applying the basic rule of differentiation which is $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, but here as it is not just the variable $x$, but rather a function of $x$, which is \[\dfrac{d}{{dx}}\left( {{f^n}(x)} \right) = n{f^{n - 1}}(x)\dfrac{d}{{dx}}\left( {f(x)} \right)\], now applying this below as shown:
\[ \Rightarrow \dfrac{d}{{dx}}f(x) = 3{\sin ^2}\left( {2x + 1} \right)\dfrac{d}{{dx}}\left( {2x + 1} \right)\]
Now simplifying the above expression as shown below:
\[ \Rightarrow \dfrac{d}{{dx}}f(x) = 3{\sin ^2}\left( {2x + 1} \right)\left( 2 \right)\]
\[ \Rightarrow \dfrac{d}{{dx}}f(x) = 6{\sin ^2}\left( {2x + 1} \right)\]
Hence the differentiation of the given expression $y = {\sin ^3}\left( {2x + 1} \right)$ using the chain rule is equal to, which is given by: \[6{\sin ^2}\left( {2x + 1} \right)\].
\[\therefore \dfrac{{dy}}{{dx}} = 6{\sin ^2}\left( {2x + 1} \right)\]

The differentiation of the given expression $y = {\sin ^3}\left( {2x + 1} \right)$ using the chain rule is equal to \[6{\sin ^2}\left( {2x + 1} \right)\].

Note: Please note that there are basic differentiation rules in chain rule of differentiation. The sum rule says the derivative of a sum of functions is the sum of their derivatives. The difference rule says the derivative of a difference of functions is the difference of their derivatives. Also remember the basic derivatives such as:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cot } \right) = - \cos e{c^2}x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {cosecx} \right) = - cosecx\cot x$