Question

If \( y = \log_e \left[ e^{3x} \left( \frac{x - 4}{x + 3} \right)^{3/2} \right] \), then find \( \frac{dy}{dx} \):

• A. \( 3 + \frac{21}{2(x - 4)(x + 3)} \)
• B. \( 3 + \frac{21}{(x - 4)(x + 3)} \)
• C. \( 3 + \frac{21}{2(x + 3)(x - 4)} \)
• D. \( 3 + \frac{7}{(x - 4)(x + 3)} \)

Hint:

When differentiating logarithmic functions, first simplify the expression using logarithmic properties, then apply the chain rule to differentiate.

Answer 1

The Correct Option is A

We have the function: \[ y = \log_e \left[ e^{3x} \left( \frac{x - 4}{x + 3} \right)^{3/2} \right]. \] Step 1: Simplify the logarithmic expression using properties of logarithms. \[ y = \log_e \left( e^{3x} \right) + \log_e \left( \left( \frac{x - 4}{x + 3} \right)^{3/2} \right). \] Using the property of logarithms \( \log_e(a^b) = b \log_e(a) \), we have: \[ y = 3x + \frac{3}{2} \log_e \left( \frac{x - 4}{x + 3} \right). \] Step 2: Differentiate \( y \) with respect to \( x \). The derivative of \( 3x \) is 3, and for the second term, apply the chain rule to differentiate the logarithmic expression: \[ \frac{dy}{dx} = 3 + \frac{3}{2} \cdot \frac{d}{dx} \left[ \log_e \left( \frac{x - 4}{x + 3} \right) \right]. \] Now, use the derivative of a logarithmic function \( \frac{d}{dx} \left[ \log_e \left( \frac{u}{v} \right) \right] = \frac{1}{u/v} \cdot \frac{d}{dx} \left( \frac{u}{v} \right) \): \[ \frac{d}{dx} \left[ \log_e \left( \frac{x - 4}{x + 3} \right) \right] = \frac{1}{\frac{x - 4}{x + 3}} \cdot \frac{d}{dx} \left[ \frac{x - 4}{x + 3} \right]. \] Then differentiate \( \frac{x - 4}{x + 3} \) using quotient rule: \[ \frac{d}{dx} \left[ \frac{x - 4}{x + 3} \right] = \frac{(x + 3)(1) - (x - 4)(1)}{(x + 3)^2} = \frac{7}{(x + 3)^2}. \] Substitute this into derivative expression: \[ \frac{dy}{dx} = 3 + \frac{3}{2} \cdot \frac{7}{(x - 4)(x + 3)} = 3 + \frac{21}{2(x - 4)(x + 3)}. \] The correct answer is \( 3 + \frac{21}{2(x - 4)(x + 3)} \).