Question

Differentiate:
${\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)w.r.t{\cos }^{-1}\left(2x\sqrt{1-x^2}\right)$, when $x\ne 0$ and show that its second derivative is zero.

Answer 1

$\begin{array}{c}{y}_1={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\\ putx=\cos \theta \\ ={\tan }^{-1}\left(\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }\right)\\ ={\tan }^{-1}\left(\tan \theta \right)\\ \Rightarrow y_1={\cos }^{-1}x\\ \Rightarrow \frac{d{y}_1}{dx}=-\frac{1}{\sqrt{1-x^2}}\\ y_2={\cos }^{-1}\left(2x\sqrt{1-x^2}\right)\\ putx=\cos \theta \\ ={\cos }^{-1}\left(2\cos \theta .\sin \theta \right)\\ ={\cos }^{-1}\left(\sin 2\theta \right)\\ ={\cos }^{-1}\left(\cos \left\{\frac{\pi }{2}-2\theta \right\}\right)\\ \Rightarrow y_2=\frac{\pi }{2}-2{\cos }^{-1}x\\ \Rightarrow \frac{d{y}_2}{dx}=\frac{2}{\sqrt{1-x^2}}\\ \therefore \frac{d{y}_1}{dx}\frac{dx}{d{y}_2}=-\frac{1}{\sqrt{1-x^2}}.\frac{\sqrt{1-x^2}}{2}\\ \Rightarrow \frac{d{y}_1}{d{y}_2}=-\frac{1}{2}\end{array}$

hence, it is clearly seen that its second derivative should be zero.