Mathematics - Differentiation Question with Solution | TestHub

Let $x^2=\cos (2\theta )$. Since $x\in (0,1)$, $x^2\in (0,1)$, which implies $2\theta \in (0,\pi /2)$, so $\theta \in (0,\pi /4)$. Using the half-angle identities:$\sqrt{1+x^2}=\sqrt{1+\cos (2\theta )}=\sqrt{2{\cos }^2\theta }=\sqrt{2}|\cos \theta |$. Since $\theta \in (0,\pi /4)$, $\cos \theta >0$, so $\sqrt{1+x^2}=\sqrt{2}\cos \theta$.$\sqrt{1-x^2}=\sqrt{1-\cos (2\theta )}=\sqrt{2{\sin }^2\theta }=\sqrt{2}|\sin \theta |$. Since $\theta \in (0,\pi /4)$, $\sin \theta >0$, so $\sqrt{1-x^2}=\sqrt{2}\sin \theta$. Substitute these into the expression for $y$:$$y={\tan }^{-1}\left(\frac{\sqrt{2}\cos \theta +\sqrt{2}\sin \theta }{\sqrt{2}\cos \theta -\sqrt{2}\sin \theta }\right)={\tan }^{-1}\left(\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\right)$$ Divide the numerator and denominator by $\cos \theta$:$$y={\tan }^{-1}\left(\frac{1+\tan \theta }{1-\tan \theta }\right)={\tan }^{-1}\left(\tan \left(\frac{\pi }{4}+\theta \right)\right)$$ Since $\theta \in (0,\pi /4)$, we have $\frac{\pi }{4}+\theta \in (\pi /4,\pi /2)$. In this interval, ${\tan }^{-1}(\tan A)=A$. So, $y=\frac{\pi }{4}+\theta$. From $x^2=\cos (2\theta )$, we have $2\theta ={\cos }^{-1}(x^2)$, so $\theta =\frac{1}{2}{\cos }^{-1}(x^2)$. Therefore, $y=\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}(x^2)$. Now, differentiate $y$ with respect to $x$:$$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{\pi }{4}\right)+\frac{d}{dx}\left(\frac{1}{2}{\cos }^{-1}(x^2)\right)$$ $$\frac{dy}{dx}=0+\frac{1}{2}\cdot \left(\frac{-1}{\sqrt{1-(x^2{)}^2}}\right)\cdot (2x)$$ $$\frac{dy}{dx}=\frac{-x}{\sqrt{1-x^4}}$$.