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You are asked to find the derivative of the following function:

\[
y = \frac{a \sin x + b \cos x}{a \sin x - b \cos x}
\]

where \( a \) and \( b \) are nonzero constants. This is a quotient of two trigonometric functions, so we will apply the **quotient rule** for differentiation.

### Quotient Rule
The quotient rule states that if you have a function of the form \( y = \frac{u(x)}{v(x)} \), its derivative is:

\[
\frac{dy}{dx} = \frac{v(x) \cdot \frac{du}{dx} - u(x) \cdot \frac{dv}{dx}}{[v(x)]^2}
\]

#### Step 1: Define \( u(x) \) and \( v(x) \)
- \( u(x) = a \sin x + b \cos x \)
- \( v(x) = a \sin x - b \cos x \)

#### Step 2: Compute derivatives of \( u(x) \) and \( v(x) \)
- \( \frac{du}{dx} = a \cos x - b \sin x \) (by differentiating \( \sin x \) and \( \cos x \))
- \( \frac{dv}{dx} = a \cos x + b \sin x \) (same process for \( v(x) \))

#### Step 3: Apply the quotient rule
Using the quotient rule formula, we get:

\[
\frac{dy}{dx} = \frac{(a \sin x - b \cos x) \cdot (a \cos x - b \sin x) - (a \sin x + b \cos x) \cdot (a \cos x + b \sin x)}{(a \sin x - b \cos x)^2}
\]

#### Step 4: Simplify the expression
Expand both the numerator and denominator:

- First term in the numerator: \( (a \sin x - b \cos x)(a \cos x - b \sin x) = a^2 \sin x \cos x - ab \sin^2 x - ab \cos^2 x + b^2 \sin x \cos x \)
- Second term in the numerator: \( (a \sin x + b \cos x)(a \cos x + b \sin x) = a^2 \sin x \cos x + ab \sin^2 x + ab \cos^2 x + b^2 \sin x \cos x \)

Now subtract the two expanded terms:

\[
\frac{dy}{dx} = \frac{\left( a^2 \sin x \cos x - ab \sin^2 x - ab \cos^2 x + b^2 \sin x \cos x \right) - \left( a^2 \sin x \cos x + ab \sin^2 x + ab \cos^2 x + b^2 \sin x \cos x \right)}{(a \sin x - b \cos x)^2}
\]

Most terms cancel out, and you're left with:

\[
\frac{dy}{dx} = \frac{-2ab (\sin^2 x + \cos^2 x)}{(a \sin x - b \cos x)^2}
\]

Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), the expression simplifies to:

\[
\frac{dy}{dx} = \frac{-2ab}{(a \sin x - b \cos x)^2}
\]

### Final Answer:
\[
\boxed{\frac{dy}{dx} = \frac{-2ab}{(a \sin x - b \cos x)^2}}
\]

Would you like more details or have any further questions?

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Here are 5 related questions you might find interesting:
1. What happens if \( a = b \)? How does the derivative change?
2. How can the quotient rule be generalized to more complex trigonometric functions?
3. Can you find the second derivative of this function?
4. How would the derivative change if the trigonometric functions were replaced with hyperbolic functions?
5. How can you apply this process to solve real-world physics problems involving waveforms?

**Tip:** Always look for symmetry or trigonometric identities like \( \sin^2 x + \cos^2 x = 1 \) when simplifying expressions involving trigonometric functions.

Find the derivative of the following function, where a and b are nonzero constants: y = (a sin x + b cos x) / (a sin x - b cos x)

Find the derivative of y = (a sin x + b cos x) / (a sin x - b cos x) using the quotient rule. This solution explains key calculus concepts like trigonometric differentiation, the quotient rule, and the Pythagorean identity, step-by-step. Ideal for advanced high school or college calculus students.

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