Question

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

Answer 1

$$\text{ Let, y }={\cos }^{-1}\left\{\frac{\cos x+\sin x}{\sqrt{2}}\right\}$$

$$y={\cos }^{-1}\{\left(\frac{1}{\sqrt{2}}\right)\cos x+\left(\frac{1}{\sqrt{2}}\right)\sin x\}$$

$$y={\cos }^{-1}\{\cos \frac{\pi }{4}\cos x+\sin \frac{\pi }{4}\sin x\}$$

$$y={\cos }^{-1}\{\cos (\frac{\pi }{4}-x)\}...\left(i\right)$$

$$\text{ Here },-\frac{\pi }{4}<x<\frac{\pi }{4}$$

$$\Rightarrow \frac{\pi }{4}>-x>-\frac{\pi }{4}$$

$$\Rightarrow -\frac{\pi }{4}<-x<\frac{\pi }{4}$$

$$\Rightarrow (-\frac{\pi }{4}+\frac{\pi }{4})<(-x+\frac{\pi }{4})<(\frac{\pi }{4}+\frac{\pi }{4})$$

$$\Rightarrow 0<(\frac{\pi }{4}-x)<\frac{\pi }{2}$$

$$\text{ So, from equation }\left(i\right),$$

$$y=\frac{\pi }{4}-x[\text{ Since },{\cos }^{-1}(\cos \theta )=\theta ,\text{ if }\theta \in [0,\pi ]]$$

$$\text{ Differentiating it with respect to x },$$

$$\frac{dy}{dx}=0-1$$

$$\frac{dy}{dx}=-1$$