Question

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

Answer 1

Hint: We are given the expression. We have to find the derivative of the function. First, apply the differentiation with respect to x on both sides of the equation. Then, apply the chain rule of differentiation to the cosecant function, and then to the angle of cosecant function.

Complete step by step solution:
We are given the expression, $y = {\csc ^3}\left( {2x} \right)$.

Differentiate both sides with respect to x.

$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\csc ^3}\left( {2x} \right)$

Now, apply the chain rule of derivatives to the right hand side of the function

$ \Rightarrow \dfrac{{dy}}{{dx}} = 3\left( {{{\csc }^2}\left( {2x} \right)} \right) \times \dfrac{d}{{dx}}\left( {\csc \left( {2x} \right)} \right)$

Now, find the derivative of the expression $\dfrac{d}{{dx}}\left( {\csc \left( {2x} \right)} \right)$ using the chain rule.

$ \Rightarrow \dfrac{d}{{dx}}\left( {\csc \left( {2x} \right)} \right) = - \csc \left( {2x} \right)\cot \left( {2x} \right)\dfrac{d}{{dx}}\left( {2x} \right)$

Again determine the derivative of the function by applying the power rule.

$ \Rightarrow \dfrac{d}{{dx}}\left( {\csc \left( {2x} \right)} \right) = - 2\csc \left( {2x} \right)\cot \left( {2x} \right)$

Now, write the derivative of the function.

$ \Rightarrow \dfrac{{dy}}{{dx}} = 3\left( {{{\csc }^2}\left( {2x} \right)} \right) \times \left( { - 2\csc \left( {2x} \right)\cot \left( {2x} \right)} \right)$

$ \Rightarrow \dfrac{{dy}}{{dx}} = - 6{\csc ^2}\left( {2x} \right)\csc \left( {2x} \right)\cot \left( {2x} \right)$

Now, apply the product rule of exponents.

$ \Rightarrow \dfrac{{dy}}{{dx}} = - 6{\csc ^3}\left( {2x} \right)\cot \left( {2x} \right)$

Hence, the derivative of the expression $y = {\csc ^3}\left( {2x} \right)$ is $ - 6{\csc ^3}\left( {2x} \right)\cot \left( {2x} \right)$.

Note: The students must note that in such type of questions, the chain rule of derivatives must be applied which is given by $\dfrac{{dy}}{{dx}} = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$. The chain rule is applied in the questions where the functions are composite functions. The composite function is the function which is made by polynomial function along with trigonometric function.
The common formulas of differentiation are:
The power rule of differentiation is ${\left( {{x^n}} \right)^\prime } = n{x^{n - 1}}$
$ \Rightarrow \dfrac{d}{{dx}}\csc x = - \csc x\cot x$
$ \Rightarrow \dfrac{d}{{dx}}\sin x = \cos x$
$ \Rightarrow \dfrac{d}{{dx}}\cos x = - \sin x$
$ \Rightarrow \dfrac{d}{{dx}}\tan x = {\sec ^2}x$
$ \Rightarrow \dfrac{d}{{dx}}\cot x = - {\csc ^2}x$