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

Let u= ^ -1 (2ax 1-a^2x^2 ) Put ax = = ^ -1 (ax) = ^ -1 (2 1- ^ 2 ) = ^ -1 (2 ) = ^ -1 ( 2 ) .....(i) And Let, v= 1-a^2x^2 Differentiating it with respect to x, dv dx =1 2 1-a^2x^2 d dx (1-a^2x^2 ) dv dx = ( 0-2a^2x 2 1-a^2x^2 ) dv dx = -a^2x 1-a^2x^2 .....(ii) Here, -1 2 <ax<1 2 -1 2 < <1 2 - 4 < < 4 So, from equation (i), u=2 [Since, ^ -1 ( )= ,if [- 2 , 2 ] ] =2 ^ -1 x Differentiating it with respect to x, du dx =2 1 1-(ax)^2 d dx (ax) du dx =2 1-a^2x^2 (a) du dx = 2a 1-a^2x^2 .....(iii) Dividing equation (iii) by (ii), du dx dv dx = ( 2a 1-a^2x^2 ) ( 1-a^2x^2 -a^2x ) du dv =-2 ax

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

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Question

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

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Answer

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