Differentiate the function $\cos^{-1} \left( \frac{\sin x + \cos x}{\sqrt{2}} \right), \quad -\frac{\pi}{4} < x < \frac{\pi}{4}$ with respect to $x$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $-1$ ##
Let $y = \cos^{-1} \left( \frac{\sin x + \cos x}{\sqrt{2}} \right)$
$\Rightarrow y = \cos^{-1} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right)$
$\Rightarrow y = \cos^{-1} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \right)$
$\Rightarrow y = \cos^{-1} \left( \cos \frac{\pi}{4} \cos x + \sin \frac{\pi}{4} \sin x \right) \quad \left[ ∵\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} \right]$
$\Rightarrow y = \cos^{-1} \left( \cos \left( \frac{\pi}{4} - x \right) \right) \quad [∵\cos(a - b) = \cos a \cos b + \sin a \sin b]$
$∴y = \frac{\pi}{4} - x$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - x \right) \Rightarrow \frac{dy}{dx} = -1$