Question

If $y={\cos }^{-1}\left\{\frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}\right\},\mathrm{find}\frac{dy}{dx}$.

Answer 1

$$\begin{gathered} \mathrm{Let}y={\cos }^{-1}\left\{\frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}\right\} \\ \mathrm{Put},x=\cos \theta \\ y={\cos }^{-1}\left\{\frac{2\cos \theta -3\sqrt{1-{\cos }^2\theta }}{\sqrt{13}}\right\} \\ y={\cos }^{-1}\left\{\frac{2\cos \theta -3\sin \theta }{\sqrt{13}}\right\} \\ y={\cos }^{-1}\left\{\cos \theta \left(\frac{2}{\sqrt{13}}\right)+\sin \theta \left(\frac{3}{\sqrt{13}}\right)\right\} \\ \mathrm{Let}\cos \varphi =\frac{2}{\sqrt{13}}\mathrm{and}\sin \varphi =\frac{3}{\sqrt{13}} \\ y={\cos }^{-1}\left\{\cos \theta \cos \varphi +\sin \theta \sin \varphi \right\} \\ y={\cos }^{-1}\left\{\cos \left(\theta -\varphi \right)\right\}...\left(\mathrm{i}\right) \\ y=\left(\theta -\varphi \right) \\ y={\cos }^{-1}x-\varphi \\ \mathrm{Differentiating}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}+0 \\ \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}} \end{gathered}$$