Question

Derivative of \( \cos^{-1} \left( \frac{\sin x + \cos x}{\sqrt{2}} \right) \), \( -\frac{\pi}{4} < x < \frac{\pi}{4} \) with respect to \( x \) is :

• A. \( -1 \)
• B. \( 1 \)
• C. \( \frac{\pi}{4} \)
• D. \( -\frac{\pi}{4} \)

Hint:

Always try to simplify the expression inside an inverse trigonometric function into the form of the corresponding direct function to "cancel" them out. Check the given range carefully to handle signs.

Answer 1

The Correct Option is A

Concept:
• Simplify the trigonometric expression inside the inverse function using the formula \( \cos(A - B) = \cos A \cos B + \sin A \sin B \).
• \( \cos^{-1}(\cos \theta) = \theta \) if \( \theta \) lies in the principal value branch \( [0, \pi] \).

Step 1: Simplify the trigonometric expression
Let \( y = \cos^{-1} \left( \frac{\sin x + \cos x}{\sqrt{2}} \right) \).
Rewrite the term inside as:
\[ \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \]
Since \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), we can write:
\[ \sin \frac{\pi}{4} \sin x + \cos \frac{\pi}{4} \cos x = \cos \left( x - \frac{\pi}{4} \right) \]
Thus, \( y = \cos^{-1} \left( \cos \left( x - \frac{\pi}{4} \right) \right) \).

Step 2: Check the range for the principal value
We are given \( -\frac{\pi}{4} < x < \frac{\pi}{4} \).
Subtract \( \frac{\pi}{4} \) from all sides:
\[ -\frac{\pi}{4} - \frac{\pi}{4} < x - \frac{\pi}{4} < \frac{\pi}{4} - \frac{\pi}{4} \]
\[ -\frac{\pi}{2} < x - \frac{\pi}{4} < 0 \]
Since \( \cos(-\theta) = \cos \theta \), we have \( \cos(x - \pi/4) = \cos(\pi/4 - x) \).
In this case, \( 0 < \frac{\pi}{4} - x < \frac{\pi}{2} \), which is inside the principal range \( [0, \pi] \).
So, \( y = \frac{\pi}{4} - x \).

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