Differentiate ^ -1 ( x 1-x^2 ) with respect to ^ -1 (2x 1-x^2 ), if -1 2 <x<1 2

Let, u= ^ -1 ( x 1-x^2 ) Put x= = ^ -1 x = ^ -1 ( 1- ^ 2 ) = ^ -1 ( ) = ^ -1 ( ) .....(i) And Let, v= ^ -1 (2x 1-x^2 ) v= ^ -1 (2 1- ^ 2 ) v- ^ -1 (2 ) v= ^ -1 ( 2 ) .....(ii) Here, -1 2 <x<1 2 -1 2 < <1 2 - 4 < < 4 So, from equation (i) u= [since, ^ -1 ( )= ,if (- 2 , 2 ) ] = ^ -1 x Differentiatiating it with respect to x, du dx =1 1-x^2 .....(iii) From equation (ii), v=2 [Since, ^ -1 ( )= ,if [- 2 , 2 ] ] =2 ^ -1 x Differentiating it with respect to x, dv dx =2 1-x^2 .....(iv) Dividing equation (iii) by (iv), du dx dv dx = (1 1-x^2 ) ( 1-x^2 2 ) du dv =1 2

Differentiate ^ -1 ( x 1-x^2 ) with respect to ^ -1 (2x 1-x^2 ), …

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Question

Differentiate $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$ with respect to $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$

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Answer

Let, $\text{u}=\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
Put $\text{x}=\sin\theta$
$\Rightarrow\theta=\sin^{-1}\text{x}$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}(\tan\theta)\ .....(\text{i})$
And
Let, $\text{v}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$
$\text{v}=\sin^{-1}(2\sin\theta\sqrt{1-\sin^{2}\theta})$
$\text{v}-\sin^{-1}(2\sin\theta\cos\theta)$
$\text{v}=\sin^{-1}(\sin2\theta)\ .....(\text{ii})$
Here,
$-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i)
$\text{u}=\theta\Big[\text{since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\sin^{-1}\text{x}$
Differentiatiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=2\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{v}=2\sin^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)\Big(\frac{\sqrt{1-\text{x}^2}}{2}\Big)$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{1}{2}$