Solveeit Logo
Login

Question

Question: If \(y = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} -...

If y=tan⁡−1(1+x2+1−x21+x2−1−x2);x2≤1y = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right);{x^2} \le 1 then find dydx\dfrac{{dy}}{{dx}} .

Explanation

Solution

In mathematics evaluating an inverse trigonometric function is same as asking what angle did, we plug into the trigonometric function to get ‘x’. The derivation of the given inverse trigonometric function is very difficult as the argument itself is very complicated. So, in order to find the derivative, we need to simplify the argument.

Complete Step by Step Solution
Given: y=tan⁡−1(1+x2+1−x21+x2−1−x2),x2<1y = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right),{x^2} < 1
First of all, since the argument of the inverse trigonometric function is very complex, we try to convert it into simple form preferably in a trigonometric form.
Let us put the value of x2=cos⁡2θ{x^2} = \cos 2\theta
Therefore, 1+x2=1+cos⁡2θ\sqrt {1 + {x^2}} = \sqrt {1 + \cos 2\theta }
We know that, 1+cos⁡2θ=2cos⁡2θ\sqrt {1 + \cos 2\theta } = \sqrt {2{{\cos }^2}\theta }
=2cos⁡θ= \sqrt 2 \cos \theta
Similarly, 1−x2=2sin⁡2θ=2sin⁡θ\sqrt {1 - {x^2}} = \sqrt {2{{\sin }^2}\theta } = \sqrt 2 \sin \theta
Thus, putting the value of 1+x2\sqrt {1 + {x^2}} as 2cos⁡θ\sqrt 2 \cos \theta and 1−x2\sqrt {1 - {x^2}} as 2sin⁡θ\sqrt 2 \sin \theta , we get
y=tan⁡−1(2cos⁡θ+2sin⁡θ2cos⁡θ−2sin⁡θ)y = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt 2 \cos \theta + \sqrt 2 \sin \theta }}{{\sqrt 2 \cos \theta - \sqrt 2 \sin \theta }}} \right)
This gives,
y=tan⁡−1(cos⁡θ+sin⁡θcos⁡θ−sin⁡θ)y = {\tan ^{ - 1}}\left( {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right)
Solving (cos⁡θ+sin⁡θcos⁡θ−sin⁡θ)\left( {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right)separately where we divide in both numerator and denominator, it gives –
(cos⁡θ+sin⁡θcos⁡θ−sin⁡θ)=1+tan⁡θ1−tan⁡θ\left( {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right) = \dfrac{{1 + \tan \theta }}{{1 - \tan \theta }}
Again,
(1+tan⁡θ1−tan⁡θ)=tan⁡π4+tan⁡θtan⁡π4−tan⁡θ=tan⁡(π4+θ)\left( {\dfrac{{1 + \tan \theta }}{{1 - \tan \theta }}} \right) = \dfrac{{\tan \dfrac{\pi }{4} + \tan \theta }}{{\tan \dfrac{\pi }{4} - \tan \theta }} = \tan \left( {\dfrac{\pi }{4} + \theta } \right)
This gives the equation to be –
y=tan⁡−1(tan⁡(π4+θ))y = {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} + \theta } \right)} \right)
The inverse trigonometric and the trigonometric function cancel out –
⇒y=π4+θ\Rightarrow y = \dfrac{\pi }{4} + \theta
Substituting the value of θ=cos⁡−1x22\theta = \dfrac{{{{\cos }^{ - 1}}{x^2}}}{2} [ as cos⁡2θ=x2\cos 2\theta = {x^2}]
We get –
⇒y=π4+cos⁡−1x22\Rightarrow y = \dfrac{\pi }{4} + \dfrac{{{{\cos }^{ - 1}}{x^2}}}{2}
Differentiating both sided with respect to x, we get –
⇒dydx=0+(−11−(x2)2×2x×12)\Rightarrow \dfrac{{dy}}{{dx}} = 0 + \left( {\dfrac{{ - 1}}{{{{\sqrt {1 - \left( {{x^2}} \right)} }^2}}} \times 2x \times \dfrac{1}{2}} \right) [ since ddxcos⁡−1x=−11−x2\dfrac{d}{{dx}}{\cos ^{ - 1}}x = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }} ]
Therefore,
⇒dydx=−x1−x4\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{{\sqrt {1 - {x^4}} }}
So, the differentiation of y=tan⁡−1(1+x2+1−x21+x2−1−x2)y = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right) gives −x1−x4\dfrac{{ - x}}{{\sqrt {1 - {x^4}} }}

Note:
The student should be able to use methods of substitutions to reduce calculation. We must remember the basic trigonometric formulas as they are very useful in reducing the complex inverse trigonometric functions into simple expressions.