Solveeit Logo
Login

Question

Question: If \(y = {\cot ^{ - 1}}\left( {\dfrac{{1 - 3{x^2}}}{{3x - {x^3}}}} \right)\); \(\left| x \right| > \...

If y=cot⁡−1(1−3x23x−x3)y = {\cot ^{ - 1}}\left( {\dfrac{{1 - 3{x^2}}}{{3x - {x^3}}}} \right); ∣x∣>13\left| x \right| > \dfrac{1}{{\sqrt 3 }} then evaluate dydx\dfrac{{dy}}{{dx}}

Explanation

Solution

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(1−3x23x−x3)y = {\cot ^{ - 1}}\left( {\dfrac{{1 - 3{x^2}}}{{3x - {x^3}}}} \right) ; where ∣x∣>13\left| x \right| > \dfrac{1}{{\sqrt 3 }}.
Here we will first simplify the given equation.
Let x=cot⁡θx = \cot \theta
So we can write it as cot⁡−1x=θ{\cot ^{ - 1}}x = \theta ………………….(1)\left( 1 \right)
Now, we will substitute the value x=cot⁡θx = \cot \theta in the given equation. Therefore, we get
⇒y=cot⁡−1(1−3cot⁡2θ3cot⁡θ−cot⁡3θ)\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(x)=tan⁡−1(1x){\cot ^{ - 1}}\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right).
Using this identity in above equation, we get
⇒y=tan⁡−1(3cot⁡θ−cot⁡3θ1−3cot⁡2θ)\Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{3\cot \theta - {{\cot }^3}\theta }}{{1 - 3{{\cot }^2}\theta }}} \right)
On multiplying −1 - 1 to numerator and denominator, we get
⇒y=tan⁡−1(cot⁡3θ−3cot⁡θ3cot⁡2θ−1)\Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{{{\cot }^3}\theta - 3\cot \theta }}{{3{{\cot }^2}\theta - 1}}} \right)
Using trigonometric identities cot⁡3θ=cot⁡3θ−3cot⁡θ3cot⁡2θ−1\cot 3\theta = \dfrac{{{{\cot }^3}\theta - 3\cot \theta }}{{3{{\cot }^2}\theta - 1}} in the above equation, we get
⇒y=tan⁡−1(cot⁡3θ)\Rightarrow y = {\tan ^{ - 1}}\left( {\cot 3\theta } \right)
We know that periodic trigonometric identities cot⁡θ=tan⁡(π2−θ)\cot \theta = \tan \left( {\dfrac{\pi }{2} - \theta } \right).
Now, we will use this trigonometric identity in the above equation. Therefore, we get
⇒y=tan⁡−1(tan⁡(π2−3θ))\Rightarrow y = {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{2} - 3\theta } \right)} \right)
Now, using the inverse trigonometric identity tan⁡−1(tan⁡θ)=θ{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta , we get
⇒y=π2−3θ\Rightarrow y = \dfrac{\pi }{2} - 3\theta
Now, we will substitute the value of θ\theta as obtained in equation (1)\left( 1 \right) in the above equation. Therefore, we get
⇒y=π2−3cot⁡−1x\Rightarrow y = \dfrac{\pi }{2} - 3{\cot ^{ - 1}}x
Now, we will differentiate both sides of the equation with respect to xx.
⇒dydx=ddx(π2−3cot⁡−1x)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{\pi }{2} - 3{{\cot }^{ - 1}}x} \right)
⇒dydx=0−3(−11+x2)\Rightarrow \dfrac{{dy}}{{dx}} = 0 - 3\left( { - \dfrac{1}{{1 + {x^2}}}} \right)
On further simplification, we get
⇒dydx=31+x2\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.