The given function is:

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

We need to find the derivative $\frac{dy}{dx}$. To do this, we'll use the quotient rule for derivatives, which is given by:

$\frac{dy}{dx} = \frac{(u'v - uv')}{v^2}$

Where:

• $u = a \sin x + b \cos x$ (the numerator)
• $v = a \sin x - b \cos x$ (the denominator)

Step 1: Derivatives of $u$ and $v$

• $u = a \sin x + b \cos x$, so: $u' = a \cos x - b \sin x$
• $v = a \sin x - b \cos x$, so: $v' = a \cos x + b \sin x$

Step 2: Apply the quotient rule

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

Step 3: Simplify the numerator

Expand both terms in the numerator.

• First term: $(a \cos x - b \sin x)(a \sin x - b \cos x)$ $= a^2 \cos x \sin x - ab \cos^2 x - ab \sin^2 x + b^2 \sin x \cos x$

• Second term: $(a \sin x + b \cos x)(a \cos x + b \sin x)$ $= a^2 \sin x \cos x + ab \sin^2 x + ab \cos^2 x + b^2 \cos x \sin x$

Now subtract the second term from the first: $(a^2 \cos x \sin x - ab \cos^2 x - ab \sin^2 x + b^2 \sin x \cos x) - (a^2 \sin x \cos x + ab \sin^2 x + ab \cos^2 x + b^2 \cos x \sin x)$

After canceling like terms and simplifying, you will find that all terms cancel out, and the result is: $\frac{dy}{dx} = 0$

Thus, the derivative of the given function is:

$\frac{dy}{dx} = 0$

Would you like further details or have any questions?

Here are some related questions:

1. What is the quotient rule and how does it apply to complex functions?
2. How do we simplify trigonometric identities when differentiating functions?
3. Can you explain why the derivative of this function is 0?
4. How do constants affect the behavior of trigonometric functions when differentiating?
5. What are some examples of other functions where the derivative simplifies to 0?

Tip: When differentiating quotients, always check if terms cancel out in the numerator before continuing the simplification.