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:
We are asked to find the derivative of a curve at a specific point \(P\).
First, we must find the full coordinates \((x, y)\) of the point \(P\) by substituting the given \(y\)-value into the curve's equation.
Step 2: Key Formula or Approach:
Equate \(4\log_e(2) = \log_e(x^3 + 24)\) and solve for \(x\) using \(\log(a^b) = b\log(a)\).
Find the derivative \(\frac{dy}{dx}\) using the chain rule: \(\frac{d}{dx}(\log_e(u)) = \frac{1}{u} \cdot \frac{du}{dx}\).
Evaluate \(\frac{dy}{dx}\) at the found \(x\).
Step 3: Detailed Explanation:
Find the \(x\)-coordinate of \(P\):
\[ y = 4\log_e(2) = \log_e(2^4) = \log_e(16) \] Equate this to the curve's equation:
\[ \log_e(16) = \log_e(x^3 + 24) \] Since the logarithm function is one-to-one:
\[ 16 = x^3 + 24 \] \[ x^3 = 16 - 24 = -8 \] \[ x = -2 \] So, point \(P\) is \((-2, 4\log_e(2))\).
Now, find the derivative \(\frac{dy}{dx}\):
\[ y = \log_e(x^3 + 24) \] \[ \frac{dy}{dx} = \frac{1}{x^3 + 24} \cdot \frac{d}{dx}(x^3 + 24) = \frac{3x^2}{x^3 + 24} \] Evaluate 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 4: Final Answer:
The value of \(\frac{dy}{dx}\) is \(\frac{3}{4}\).