To find the derivative $f'(x)$ of the function

$f(x) = \frac{9\sin x}{2\sin x + 4\cos x},$

we use the quotient rule:

$\frac{d}{dx} \left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}.$

Here:

• $u(x) = 9\sin x$,
• $v(x) = 2\sin x + 4\cos x$.

Step 1: Compute $u'(x)$

$u(x) = 9\sin x \implies u'(x) = 9\cos x.$

Step 2: Compute $v'(x)$

$v(x) = 2\sin x + 4\cos x \implies v'(x) = 2\cos x - 4\sin x.$

Step 3: Apply the quotient rule

Substitute into the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}.$

Substitute $u(x), u'(x), v(x), v'(x)$:

[ f'(x) = \frac{[9\cos x](2\sin x + 4\cos x) - [9\sin x](2\cos x - 4\sin x)}{(2\sin x + 4\cos x)^2}. ]

Step 4: Simplify the numerator

Expand both terms in the numerator:

1. Expand $[9\cos x](2\sin x + 4\cos x)$: $9\cos x \cdot 2\sin x + 9\cos x \cdot 4\cos x = 18\cos x \sin x + 36\cos^2 x.$

2. Expand $[9\sin x](2\cos x - 4\sin x)$: $9\sin x \cdot 2\cos x - 9\sin x \cdot 4\sin x = 18\sin x \cos x - 36\sin^2 x.$

Now subtract these two terms: $(18\cos x \sin x + 36\cos^2 x) - (18\sin x \cos x - 36\sin^2 x).$

Combine like terms: $18\cos x \sin x - 18\sin x \cos x + 36\cos^2 x + 36\sin^2 x.$

Simplify further using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$: $36(\sin^2 x + \cos^2 x) = 36.$

So the numerator simplifies to: $36.$

Step 5: Write the final derivative

$f'(x) = \frac{36}{(2\sin x + 4\cos x)^2}.$

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Final Answer:

$f'(x) = \frac{36}{(2\sin x + 4\cos x)^2}.$

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Would you like a detailed example for applying this derivative? Here are some related questions to practice:

1. What is the value of $f'(x)$ at $x = \frac{\pi}{4}$?
2. How does $f(x)$ behave as $x \to 0$?
3. How do you interpret $f'(x)$ in terms of rates of change?
4. What are the critical points of $f(x)$?
5. How does the quotient rule differ from the product rule?

Tip: Always check for simplifications after using the quotient rule to avoid unnecessary complexity!