Question

Differentiate

${\cos }^{-1}\left\{\frac{\cos x+\sin x}{\sqrt{2}}\right\},-\frac{\pi }{4}<x<\frac{\pi }{4}$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\cos }^{-1}\left\{\frac{\cos x+\sin x}{\sqrt{2}}\right\} \\ y={\cos }^{-1}\left\{\left(\frac{1}{\sqrt{2}}\right)\cos x+\left(\frac{1}{\sqrt{2}}\right)\sin x\right\} \\ y={\cos }^{-1}\left\{\cos \frac{\mathrm{\pi }}{4}\cos x+\sin \frac{\mathrm{\pi }}{4}\sin x\right\} \\ y={\cos }^{-1}\left\{\cos \left(\frac{\mathrm{\pi }}{4}-x\right)\right\}...\left(\mathrm{i}\right) \\ \mathrm{Here},-\frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{\mathrm{\pi }}{4}>-x>-\frac{\mathrm{\pi }}{4} \\ \Rightarrow -\frac{\mathrm{\pi }}{4}<-x<\frac{\mathrm{\pi }}{4} \\ \Rightarrow \left(-\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{4}\right)<\left(-x+\frac{\mathrm{\pi }}{4}\right)<\left(\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow 0<\left(\frac{\mathrm{\pi }}{4}-x\right)<\frac{\mathrm{\pi }}{2} \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ y=\frac{\mathrm{\pi }}{4}-x\left[\mathrm{Since},{\cos }^{-1}\left(\cos \theta \right)=\theta ,\mathrm{if}\theta \in \left[0,\mathrm{\pi }\right]\right] \\ \mathrm{Differentiating}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dy}{dx}=0-1 \\ \frac{dy}{dx}=-1 \end{gathered}$$