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: $$\boxed{-\left( \frac{15\cos x + 6}{12\sin x - 2} \right)}.$$