Question

If \[ f(x) = 5 \cos^3 x - 3 \sin^3 x \quad \text{and} \quad g(x) = 4 \sin^3 x + \cos^2 x, \] then the derivative of \( f(x) \) with respect to \( g(x) \) is:

• A. \( \frac{5 \cos x + 2}{6 \cos x - 1} \)
• B. \( \frac{- (5 \cos x + 2)}{6 \cos x - 1} \)
• C. \( \frac{15 \cos x - 6}{12 \sin x + 2} \)
• D. \( \frac{- (15 \cos x + 6)}{12 \sin x - 2} \)

Hint:

To differentiate one function with respect to another, compute their derivatives separately and take the ratio \( \frac{f'(x)}{g'(x)} \).

Answer 1

The Correct Option is D

Step 1: Compute \( f'(x) \)
Given: \[ f(x) = 5\cos^3x - 3\sin^2x. \] Differentiating: \[ f'(x) = 5 \cdot 3\cos^2x(-\sin x) - 3 \cdot 2\sin x\cos x. \] \[ = -15\cos^2x\sin x - 6\sin x\cos x. \] Factoring: \[ f'(x) = -\sin x(15\cos^2x + 6\cos x). \] Step 2: Compute \( g'(x) \)
Given: \[ g(x) = 4\sin^3x + \cos^2x. \] Differentiating: \[ g'(x) = 4 \cdot 3\sin^2x\cos x - 2\cos x\sin x. \] \[ = 12\sin^2x\cos x - 2\cos x\sin x. \] Factoring: \[ g'(x) = \cos x(12\sin^2x - 2). \] Step 3: Compute \( \frac{df}{dg} \)
\[ \frac{df}{dg} = \frac{f'(x)}{g'(x)} = \frac{-\sin x(15\cos^2x + 6\cos x)}{\cos x(12\sin^2x - 2)}. \] Simplifying: \[ = -\frac{15\cos^2x + 6\cos x}{12\sin^2x - 2}. \] Step 4: Conclusion
Thus, the final answer is: \[ \boxed{-\left( \frac{15\cos x + 6}{12\sin x - 2} \right)}. \]