Question

If $y = {\cot ^{ - 1}}\left( {\dfrac{{1 - 3{x^2}}}{{3x - {x^3}}}} \right)$; $\left| x \right| > \dfrac{1}{{\sqrt 3 }}$ then evaluate $\dfrac{{dy}}{{dx}}$

Answer 1

Hint: Here we will first simplify the given equation using suitable trigonometric identities and also different inverse trigonometric identities. Then we will differentiate the simplified equation to get the required answer.

Complete step-by-step answer:
The given equation is $y = {\cot ^{ - 1}}\left( {\dfrac{{1 - 3{x^2}}}{{3x - {x^3}}}} \right)$ ; where $\left| x \right| > \dfrac{1}{{\sqrt 3 }}$.
Here we will first simplify the given equation.
Let $x = \cot \theta $
So we can write it as ${\cot ^{ - 1}}x = \theta $………………….$\left( 1 \right)$
Now, we will substitute the value $x = \cot \theta $ in the given equation. Therefore, we get
 $ \Rightarrow y = {\cot ^{ - 1}}\left( {\dfrac{{1 - 3{{\cot }^2}\theta }}{{3\cot \theta - {{\cot }^3}\theta }}} \right)$
We know from the inverse trigonometric identities that ${\cot ^{ - 1}}\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right)$.
Using this identity in above equation, we get
$ \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{3\cot \theta - {{\cot }^3}\theta }}{{1 - 3{{\cot }^2}\theta }}} \right)$
On multiplying $ - 1$ to numerator and denominator, we get
$ \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{{{\cot }^3}\theta - 3\cot \theta }}{{3{{\cot }^2}\theta - 1}}} \right)$
Using trigonometric identities $\cot 3\theta = \dfrac{{{{\cot }^3}\theta - 3\cot \theta }}{{3{{\cot }^2}\theta - 1}}$ in the above equation, we get
$ \Rightarrow y = {\tan ^{ - 1}}\left( {\cot 3\theta } \right)$
We know that periodic trigonometric identities $\cot \theta = \tan \left( {\dfrac{\pi }{2} - \theta } \right)$.
Now, we will use this trigonometric identity in the above equation. Therefore, we get
$ \Rightarrow y = {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{2} - 3\theta } \right)} \right)$
Now, using the inverse trigonometric identity ${\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta $, we get
$ \Rightarrow y = \dfrac{\pi }{2} - 3\theta $
Now, we will substitute the value of $\theta $ as obtained in equation $\left( 1 \right)$ in the above equation. Therefore, we get
$ \Rightarrow y = \dfrac{\pi }{2} - 3{\cot ^{ - 1}}x$
Now, we will differentiate both sides of the equation with respect to $x$.
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{\pi }{2} - 3{{\cot }^{ - 1}}x} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = 0 - 3\left( { - \dfrac{1}{{1 + {x^2}}}} \right)$
On further simplification, we get
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{3}{{1 + {x^2}}}$
Hence, this is the required answer.

Note: Trigonometric identities are defined as the equalities which include the trigonometric functions and they are always true for every value of the variables. Inverse trigonometric identities are defined as the equalities which include the trigonometric functions and they are always true for every value of the variables. Trigonometric identities are used only when the equation contains a trigonometric function.