Question

If \[ y = \log_e \left( \frac{1 + 2x^2}{1 - 3x^2} \right), \] then \( \frac{dy}{dx} \) is:

• A. \( \frac{10x}{1 - x^2 - 6x^4} \)
• B. \( \frac{12x^3}{1 - x^2 - 6x^4} \)
• C. \( \frac{10x}{1 - 6x^4} \)
• D. \( \frac{-10x}{1 - x^2 - 6x^4} \)
• E. \( \frac{-12x^3}{1 - x^2 - 6x^4} \)

Hint:

For logarithmic functions with fractions, use the quotient rule when differentiating.

Answer 1

The Correct Option is A

Step 1: Use the differentiation rule for logarithmic functions Given: \[ y = \log_e \left( \frac{1 + 2x^2}{1 - 3x^2} \right). \] Using the differentiation rule: \[ \frac{d}{dx} \log_e (f(x)) = \frac{1}{f(x)} \cdot f'(x), \] where \[ f(x) = \frac{1 + 2x^2}{1 - 3x^2}. \] Step 2: Differentiate using the quotient rule Let \( u = 1 + 2x^2 \) and \( v = 1 - 3x^2 \). 
Differentiate \( u \) and \( v \): \[ \frac{du}{dx} = 4x, \quad \frac{dv}{dx} = -6x. \] Using the quotient rule: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot du - u \cdot dv}{v^2}. \] Substituting the values: \[ \frac{d}{dx} \left( \frac{1 + 2x^2}{1 - 3x^2} \right) = \frac{(1 - 3x^2)(4x) - (1 + 2x^2)(-6x)}{(1 - 3x^2)^2}. \] Expanding: \[ = \frac{4x - 12x^3 + 6x + 12x^3}{(1 - 3x^2)^2}. \] \[ = \frac{10x}{(1 - 3x^2)^2}. \] Step 3: Substitute back into logarithmic differentiation formula \[ \frac{dy}{dx} = \frac{10x}{1 - x^2 - 6x^4}. \] Thus, the correct answer is (A) \( \frac{10x}{1 - x^2 - 6x^4} \).