Question

Differentiate $${\sin }^{-1}\left(4x\sqrt{1-4x^2}\right)$$ with respect to $$\sqrt{1-4x^2}$$ , if $$x\in (\frac{1}{2\sqrt{2}},\frac{1}{2})$$ ?

Answer 1

$$\text{ Let, u }={\sin }^{-1}\left(4x\sqrt{1-4x^2}\right)$$
$$\text{ put }2x=\cos \theta$$
$$u={\sin }^{-1}(2\times \cos \theta \sqrt{1-{\cos }^2\theta })$$
$$\Rightarrow u={\sin }^{-1}(2\cos \theta \sin \theta )$$
$$\Rightarrow u={\sin }^{-1}(\sin 2\theta )...\left(i\right)$$
$$\text{ Let, v }=\sqrt{1-4x^2}...\left(ii\right)$$
$$\text{ Here },$$
$$x\in (\frac{1}{2\sqrt{2}},\frac{1}{2})$$
$$\Rightarrow 2x\in (\frac{1}{\sqrt{2}},1)$$
$$\Rightarrow \cos \theta \in (\frac{1}{\sqrt{2}},1)$$
$$\Rightarrow \theta \in (0,\frac{\pi }{4})$$
$$\text{ So, from equation }\left(i\right),$$
$$u=2\theta ..........[\text{Since },{\sin }^{-1}\left(sin\theta \right)=\theta ,\text{ if }\theta \in (-\frac{\pi }{2},\frac{\pi }{2})]$$
$$\Rightarrow u=2{\cos }^{-1}\left(2x\right)........[\text{ Since, }2x=\cos \theta ]$$

Differentiate it with respect to x,

$$\frac{du}{dx}=2\left(\frac{-1}{\sqrt{1-{\left(2x\right)}^2}}\right)\frac{d}{dx}\left(2x\right)$$
$$\frac{du}{dx}=\frac{-2}{\sqrt{1-4x^2}}\left(2\right)$$
$$\frac{du}{dx}=\frac{-4}{\sqrt{1-4x^2}}...\left(iii\right)$$
$$\text{ Differentiating equation }\left(ii\right)\text{ with respect to x,}$$
$$\frac{dv}{dx}=\frac{1}{2\sqrt{1-4x^2}}\frac{d}{dx}(1-4x^2)$$
$$\Rightarrow \frac{dv}{dx}=\frac{1}{2\sqrt{1-4x^2}}(-8x)$$
$$\Rightarrow \frac{dv}{dx}=\frac{-4x}{\sqrt{1-4x^2}}...\left(iv\right)$$
$$\text{ Dividing equation }\left(iii\right)\text{ by }\left(iv\right)$$
$$\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{-4}{\sqrt{1-4x^2}}\times \frac{\sqrt{1-4x^2}}{-4x}$$
$$\therefore \frac{du}{dv}=\frac{1}{x}$$