1. We are asked to find the derivative of the function $$f(x) = \frac{2 \sin(x) - 7}{9x^9 - 3}$$ using the quotient rule. 2. Recall the quotient rule formula: if $$f(x) = \frac{u(x)}{v(x)}$$, then $$f'(x) = \frac{v(x) u'(x) - u(x) v'(x)}{[v(x)]^2}$$. 3. Identify $$u(x) = 2 \sin(x) - 7$$ and $$v(x) = 9x^9 - 3$$. 4. Compute the derivatives: - $$u'(x) = 2 \cos(x)$$ because derivative of $$\sin(x)$$ is $$\cos(x)$$ and derivative of constant $$-7$$ is zero. - $$v'(x) = 81x^8$$ because derivative of $$9x^9$$ is $$9 \times 9 x^{8} = 81x^8$$ and derivative of constant $$-3$$ is zero. 5. Apply the quotient rule: $$f'(x) = \frac{(9x^9 - 3)(2\cos(x)) - (2\sin(x) - 7)(81x^8)}{(9x^9 - 3)^2}$$ 6. This is the derivative using the quotient rule without expanding or simplifying further, being careful with parentheses as requested. Final answer: $$f'(x) = \frac{(9x^9 - 3)(2\cos(x)) - (2\sin(x) - 7)(81x^8)}{(9x^9 - 3)^2}$$