We are given the function $y = \frac{\cos x + \sin x}{\sin x - \cos x}$, and we need to find its derivative with respect to $x$.

Step 1: Apply the quotient rule

The quotient rule states that if we have a function of the form:

$y = \frac{f(x)}{g(x)}$

then its derivative is given by:

$\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

In our case:

• $f(x) = \cos x + \sin x$
• $g(x) = \sin x - \cos x$

Step 2: Differentiate $f(x)$ and $g(x)$

• The derivative of $f(x) = \cos x + \sin x$ is:

$f'(x) = -\sin x + \cos x$

• The derivative of $g(x) = \sin x - \cos x$ is:

$g'(x) = \cos x + \sin x$

Step 3: Apply the quotient rule

Now we can apply the quotient rule formula:

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

Step 4: Simplify the numerator

First, expand both products in the numerator:

$(-\sin x + \cos x)(\sin x - \cos x) = -\sin^2 x + \sin x \cos x + \cos x \sin x - \cos^2 x$

which simplifies to:

$-\sin^2 x + 2 \sin x \cos x - \cos^2 x$

Now expand the second product:

$(\cos x + \sin x)(\cos x + \sin x) = \cos^2 x + 2 \sin x \cos x + \sin^2 x$

So, the numerator becomes:

$-\sin^2 x + 2 \sin x \cos x - \cos^2 x - (\cos^2 x + 2 \sin x \cos x + \sin^2 x)$

Simplifying:

$-\sin^2 x + 2 \sin x \cos x - \cos^2 x - \cos^2 x - 2 \sin x \cos x - \sin^2 x$

$= -2\sin^2 x - 2\cos^2 x$

Now, factor out the common factor of $-2$:

$= -2(\sin^2 x + \cos^2 x)$

Since $\sin^2 x + \cos^2 x = 1$, this simplifies to:

$= -2$

Step 5: Final derivative

Now substitute the simplified numerator into the derivative formula:

$\frac{dy}{dx} = \frac{-2}{(\sin x - \cos x)^2}$

Final Answer:

$\frac{dy}{dx} = \frac{-2}{(\sin x - \cos x)^2}$

Let me know if you'd like further details or have any questions!

Follow-up Questions:

1. How does the quotient rule work with trigonometric functions?
2. What happens if we apply the product rule instead of the quotient rule?
3. How can you simplify trigonometric identities to help with derivatives?
4. How do you handle derivatives of functions involving both sine and cosine?
5. Can this method be generalized to higher-order derivatives of rational trigonometric functions?

Tip: Always look for trigonometric identities like $\sin^2 x + \cos^2 x = 1$ to simplify your work!