Question

The point $P(x, y)$, where $y = 4\log_e(2)$, lies on the curve with equation $y = \log_e(x^3 + 24)$. Then the value of $\frac{dy}{dx}$ at the point $P$ is

• A. $-\frac{3}{8}$
• B. $\frac{3}{8}$
• C. $-\frac{3}{4}$
• D. $\frac{3}{4}$

Hint:

When given a specific $y$ value for a curve, find $x$ first. Most derivative problems at a point require both coordinates, or at least the $x$ value, for substitution into the derivative formula.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
First, find the $x$-coordinate of point $P$ by equating the two expressions for $y$. Then, differentiate the curve equation and evaluate the derivative at $P$.

Step 2: Detailed Explanation:
1. Find $x$ at $P$:
\[ y = \log_e(x^3 + 24) = 4\log_e(2) = \log_e(2^4) = \log_e(16) \]
\[ x^3 + 24 = 16 \implies x^3 = -8 \implies x = -2 \]
2. Differentiate the curve equation:
\[ y = \log_e(x^3 + 24) \implies \frac{dy}{dx} = \frac{1}{x^3 + 24} \cdot \frac{d}{dx}(x^3 + 24) = \frac{3x^2}{x^3 + 24} \]
3. Evaluate $\frac{dy}{dx}$ at $x = -2$:
\[ \left[ \frac{dy}{dx} \right]_{x=-2} = \frac{3(-2)^2}{(-2)^3 + 24} = \frac{3(4)}{-8 + 24} = \frac{12}{16} = \frac{3}{4} \]

Step 3: Final Answer:
The derivative at point $P$ is $3/4$.