Question

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Derivative of
\[
\(\cos^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right),\)
\quad -\frac{\pi}{4}\(<x<\)\frac{\pi}{4}
\]
with respect to \(x\) is:
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• A. \( -1 \)
• B. \( 1 \)
• C. \( \frac{\pi}{4} \)
• D. \( -\frac{\pi}{4} \)

Hint:

Always simplify inverse trig functions before differentiating to avoid using the chain rule on complex radicals.
Check the range carefully to ensure the simplified variable form is valid within the principal branch.

Answer 1

The Correct Option is A

Concept:
• Simplify the trigonometric expression inside the inverse function using compound angle formulas.
• \( \cos(A - B) = \cos A \cos B + \sin A \sin B \).
• Property: \( \cos^{-1}(\cos \theta) = \theta \) if \( \theta \in [0, \pi] \).

Step 1: Simplify the internal trigonometric expression
Let \( y = \cos^{-1} \left( \frac{\sin x + \cos x}{\sqrt{2}} \right) \). Rewrite the term: \[ \frac{\sin x + \cos x}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \] We know \( \cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2} \). \[ \frac{\sin x + \cos x}{\sqrt{2}} = \cos x \cos \left( \frac{\pi}{4} \right) + \sin x \sin \left( \frac{\pi}{4} \right) = \cos \left( x - \frac{\pi}{4} \right) \]

Step 2: Apply the range constraint to simplify the inverse cosine function
Now, \( y = \cos^{-1} \left( \cos \left( x - \frac{\pi}{4} \right) \right) \). Given \( -\frac{\pi}{4} < x < \frac{\pi}{4} \). Subtracting \( \pi/4 \): \[ -\frac{\pi}{2} < x - \frac{\pi}{4} < 0 \] Since \( \cos(-\theta) = \cos \theta \), we can write \( \cos(x - \pi/4) = \cos(\pi/4 - x) \). If \( -\frac{\pi}{2} < x - \frac{\pi}{4} < 0 \), then \( 0 < \frac{\pi}{4} - x < \frac{\pi}{2} \). This value is in the principal branch \( [0, \pi] \). So, \( y = \frac{\pi}{4} - x \).

Step 3: Differentiate with respect to \( x \)
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - x \right) \] \[ \frac{dy}{dx} = 0 - 1 = -1 \] This matches option (A).