To solve the given problem $y = \tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)$ and find $\frac{dy}{dx}$, we follow these steps:

Step 1: Let the argument of the inverse tangent be $u$:

$u = \frac{\cos x - \sin x}{\cos x + \sin x}$

Step 2: Use the derivative formula for $y = \tan^{-1}(u)$:

$\frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx}$

Step 3: Simplify $u^2$:

$u^2 = \left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)^2 = \frac{(\cos x - \sin x)^2}{(\cos x + \sin x)^2}$ Using the identity $(\cos x - \sin x)^2 = \cos^2 x + \sin^2 x - 2\cos x \sin x$ and $\cos^2 x + \sin^2 x = 1$: $u^2 = \frac{1 - 2\cos x \sin x}{(\cos x + \sin x)^2}$

Step 4: Differentiate $u$ with respect to $x$:

Using the quotient rule: $\frac{du}{dx} = \frac{(\cos x + \sin x) \cdot (\sin x - \cos x) - (\cos x - \sin x) \cdot (\sin x + \cos x)}{(\cos x + \sin x)^2}$ Simplify the numerator: $(\cos x + \sin x)(\sin x - \cos x) - (\cos x - \sin x)(\sin x + \cos x)$ Expanding both terms: $= (\cos x \sin x - \cos^2 x + \sin^2 x - \sin x \cos x) - (\cos x \sin x + \cos^2 x - \sin^2 x - \sin x \cos x)$ $= -2\cos^2 x - 2\sin^2 x$ Simplify further using $\cos^2 x + \sin^2 x = 1$: $\frac{du}{dx} = \frac{-2}{(\cos x + \sin x)^2}$

Step 5: Substitute $u^2$ and $\frac{du}{dx}$ into $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{1 + \frac{1 - 2\cos x \sin x}{(\cos x + \sin x)^2}} \cdot \frac{-2}{(\cos x + \sin x)^2}$ Simplify: $\frac{dy}{dx} = \frac{-2}{(\cos x + \sin x)^2 + (1 - 2\cos x \sin x)} = \frac{-2}{1 + 2\cos x \sin x}$

Final Answer:

$\frac{dy}{dx} = \frac{-2}{1 + 2\cos x \sin x}$

Let me know if you want detailed clarifications or additional steps! Here are some related questions:

1. What is the role of the derivative of inverse trigonometric functions in calculus?
2. How can you simplify trigonometric identities involving sum and difference?
3. What is the significance of the quotient rule in this context?
4. How does the identity $\cos^2 x + \sin^2 x = 1$ simplify expressions?
5. Can the result be verified using numerical values for $x$?

Tip: When dealing with trigonometric derivatives, always simplify intermediate expressions to avoid mistakes!