Question

If \( y = \log_e (x^3 + 24) \), find \( \frac{dy}{dx} \) at \( y = \log_e 2 \).

Hint:

For logarithmic differentiation, remember the chain rule: \( \frac{d}{dx} \log_e(u(x)) = \frac{1}{u(x)} \cdot u'(x) \).

Answer 1

Step 1: Differentiate the given expression.
The given expression is \( y = \log_e (x^3 + 24) \). We differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{x^3 + 24} \cdot \frac{d}{dx} (x^3 + 24) \]
\[ \frac{dy}{dx} = \frac{1}{x^3 + 24} \cdot 3x^2 \]

Step 2: Substitute the value of \( y = \log_e 2 \).
At \( y = \log_e 2 \), we have \( x^3 + 24 = 2 \), which gives:
\[ x^3 = 2 - 24 = -22 \quad \Rightarrow \quad x = \sqrt[3]{-22} \]

Step 3: Compute \( \frac{dy}{dx} \).
Substitute \( x = \sqrt[3]{-22} \) into the derivative:
\[ \frac{dy}{dx} = \frac{3x^2}{x^3 + 24} = \frac{3(\sqrt[3]{-22})^2}{2} \]