Question
Differentiate $\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$ with respect to $\sqrt{1-4\text{x}^2},$ if:
$\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
✓
Answer
Let $\text{u}=\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$
Put $2\text{x}=\cos\theta \text{ So},$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow2\text{x}\in\Big(-1,-\frac{1}{\sqrt{2}},\Big)$
$\Rightarrow\theta\in\Big(\frac{3\pi}{4},\pi\Big)$
So, from equation (i),
$\text{u}=\pi-2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{3\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=\pi-2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{du}}{\text{dx}}=0-2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-2}{\sqrt{1-4\text{x}^2}}(2)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\ .....(\text{vi})$
From equation (iv)
$\frac{\text{dv}}{\text{dx}}=\frac{4}{\sqrt{1-4\text{x}^2}}$
But, $\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\frac{\text{dv}}{\text{dx}}=\frac{-4(-\text{x})}{\sqrt{1-4(-\text{x})^2}}$
$\frac{\text{dv}}{\text{dx}}=\frac{4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{vii})$
Dividing equation (vi) by (vii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{4\text{x}}$
$\frac{\text{du}}{\text{dv}}=-\frac{1}{\text{x}}$
Put $2\text{x}=\cos\theta \text{ So},$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow2\text{x}\in\Big(-1,-\frac{1}{\sqrt{2}},\Big)$
$\Rightarrow\theta\in\Big(\frac{3\pi}{4},\pi\Big)$
So, from equation (i),
$\text{u}=\pi-2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{3\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=\pi-2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{du}}{\text{dx}}=0-2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-2}{\sqrt{1-4\text{x}^2}}(2)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\ .....(\text{vi})$
From equation (iv)
$\frac{\text{dv}}{\text{dx}}=\frac{4}{\sqrt{1-4\text{x}^2}}$
But, $\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\frac{\text{dv}}{\text{dx}}=\frac{-4(-\text{x})}{\sqrt{1-4(-\text{x})^2}}$
$\frac{\text{dv}}{\text{dx}}=\frac{4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{vii})$
Dividing equation (vi) by (vii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{4\text{x}}$
$\frac{\text{du}}{\text{dv}}=-\frac{1}{\text{x}}$
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