Question

If \( y = \tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right) \), \( \frac{-\pi}{2}<x<\frac{\pi}{2} \), then \( \frac{dy}{dx} \) is:

• A. \( \tan x \)
• B. \( \cos x \)
• C. \( \sin x \)
• D. \( -1 \)
• E. 0

Hint:

For derivatives of inverse trigonometric functions, remember to apply the chain rule and the quotient rule correctly for rational functions.

Answer 1

The Correct Option is D

We are given: \[ y = \tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right) \] To differentiate, we use the chain rule. 
The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \). Let: \[ u = \frac{\cos x - \sin x}{\cos x + \sin x} \] We first compute \( \frac{du}{dx} \). 
Using the quotient rule, where \( f(x) = \cos x - \sin x \) and \( g(x) = \cos x + \sin x \), we get: \[ \frac{du}{dx} = \frac{g(x) \cdot \frac{d}{dx}[f(x)] - f(x) \cdot \frac{d}{dx}[g(x)]}{g(x)^2} \] Now, compute the derivatives: \[ \frac{d}{dx} [\cos x - \sin x] = -\sin x - \cos x \] \[ \frac{d}{dx} [\cos x + \sin x] = -\sin x + \cos x \] Substitute into the quotient rule: \[ \frac{du}{dx} = \frac{(\cos x + \sin x)(-\sin x - \cos x) - (\cos x - \sin x)(-\sin x + \cos x)}{(\cos x + \sin x)^2} \] Simplifying the numerator: \[ = -(\sin x + \cos x)^2 + (\cos x - \sin x)(\cos x - \sin x) \] \[ = -(\sin^2 x + 2\sin x \cos x + \cos^2 x) + (\cos^2 x - 2\sin x \cos x + \sin^2 x) \] \[ = -1 + 1 = 0 \] Thus: \[ \frac{du}{dx} = -1 \] Now, apply the derivative of \( \tan^{-1}(u) \): \[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot (-1) \] Substitute \( u = \frac{\cos x - \sin x}{\cos x + \sin x} \), which simplifies the final result: \[ \frac{dy}{dx} = -1 \] Thus, the correct value of \( \frac{dy}{dx} \) is \( \boxed{-1} \).