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

Let u= ^ -1 (2x 1-x^2 ) Put x= = ^ -1 (2 1- ^2 ) = ^ -1 (2 ) = ^ -1 ( 2 ) .....(i) And, Let v= ^ -1 (1 1-x^2 ) = ^ -1 (1 1- ^2 ) = ^ -1 (1 ) = ^ -1 ( ) = ^ -1 (1 1 ) [Since, ^ -1 x= ^ -1 (1 x ) ] = ^ -1 ( ) .....(ii) Here, x (1 2 ,1 ) (1 2 ,1 ) ( 4 , 2 ) So, from equation (i), u=2 [Since, ^ -1 ( )= , if (- 2 , 2 ) ] Let, u=2 ^ -1 x [Since,x= ] du dx =2 (1 1-x^2 ) du dx =2 1-x^2 .....(iii) And, from equation (ii), v= [Since, ^ -1 ( )= ,if [0, ] ] = ^ -1 x [Since,x= ] dv dx =1 1-x^2 .....(iv) Dividing equation (iii) by (iv), du dx dv dx =2 1-x^2 1-x^2 1 du dv =2

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

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Question

Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\sec^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big),$ if:
$\text{x}\in\Big(\frac{1}{\sqrt{2}},1\Big)$

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Answer

Let $\text{u}=\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
And,
Let $\text{v}=\sec^{-1}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\text{v}=\sec^{-1}\Big(\frac{1}{\sqrt{1-\sin^2\theta}}\Big)$
$\Rightarrow\text{v}=\sec^{-1}\Big(\frac{1}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\sec^{-1}(\sec\theta)$
$\Rightarrow\text{v}=\cos^{-1}\bigg(\frac{1}{\frac{1}{\cos\theta}}\bigg)\ \Big[\text{Since},\sec^{-1}\text{x}=\cos^{-1}\big(\frac{1}{\text{x}}\big)\Big]$
$\Rightarrow\text{v}=\cos^{-1}(\cos\theta)\ .....(\text{ii})$
Here,
$\text{x}\in\Big(\frac{1}{\sqrt{2}},1\Big)$
$\Rightarrow\sin\theta\in\Big(\frac{1}{\sqrt{2}},1\Big)$
$\Rightarrow\theta\in\Big(\frac{\pi}{4},\frac{\pi}{2}\Big)$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
Let, $\text{u}=2\sin^{-1}\text{x}\big[\text{Since},\text{x}=\sin\theta\big]$
$\frac{\text{du}}{\text{dx}}=2\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
And, from equation (ii),
$\text{v}=\theta\ \big[\text{Since},\cos^{-1}(\cos\theta)=\theta,\text{if }\theta\in[0,\pi]\big]$
$\Rightarrow\text{v}=\sin^{-1}\text{x}\big[\text{Since},\text{x}=\sin\theta\big]$
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{1}$
$\therefore\frac{\text{du}}{\text{dv}}=2$

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