Question

If \( y = \log \left[ \tan \left( \sqrt\frac{2x - 1}{2x + 1} \right) \right] \) for \( x>0 \), then find \[ \left( \frac{dy}{dx} \right)_{x = 1}. \]

• A. \( \frac{4\sqrt{2} \log 2}{9 \sin \left( \frac{2}{\sqrt{3}} \right)} \)
• B. \( \frac{4\sqrt{2} \log 2}{9 \sin \left( \frac{\sqrt{3}}{2} \right)} \)
• C. \( \frac{4\sqrt{3} \log 2}{9 \sin \left( \frac{2}{\sqrt{3}} \right)} \)
• D. \( \frac{4\sqrt{2} \log 2}{9 \sin \left( \frac{\sqrt{3}}{2} \right)} \) \bigskip

Hint:

When differentiating logarithmic and trigonometric functions, always apply the chain rule carefully, especially when they are composed. Also, ensure that you evaluate at the specified point correctly.

Answer 1

The Correct Option is C

Step 1: We are given that \( y = \log \left[ \tan \left( \frac{2x - 1}{2x + 1} \right) \right] \). To differentiate this, let: \[ u = \frac{2x - 1}{2x + 1}. \] Thus, we can rewrite \( y \) as: \[ y = \log (\tan u). \] Now, we differentiate \( y \) with respect to \( u \): \[ \frac{dy}{du} = \frac{1}{\tan u} \cdot \sec^2 u. \] Next, we differentiate \( u = \frac{2x - 1}{2x + 1} \) using the quotient rule: \[ \frac{du}{dx} = \frac{(2x + 1)(2) - (2x - 1)(2)}{(2x + 1)^2} = \frac{4}{(2x + 1)^2}. \] Thus, the total derivative of \( y \) with respect to \( x \) is: \[ \frac{dy}{dx} = \frac{1}{\tan u} \cdot \sec^2 u \cdot \frac{du}{dx}. \] \bigskip Step 2: Now, we evaluate the derivative at \( x = 1 \). First, we calculate \( u \) when \( x = 1 \): \[ u = \frac{2(1) - 1}{2(1) + 1} = \frac{1}{3}. \] Next, we find \( \frac{dy}{dx} \) at \( x = 1 \): \[ \frac{dy}{dx} = \frac{1}{\tan \left( \frac{1}{3} \right)} \cdot \sec^2 \left( \frac{1}{3} \right) \cdot \frac{4}{(2(1) + 1)^2} = \frac{1}{\tan \left( \frac{1}{3} \right)} \cdot \sec^2 \left( \frac{1}{3} \right) \cdot \frac{4}{9}. \] \bigskip Step 3: Simplifying the expression gives the final result: \[ \frac{4\sqrt{3} \log 2}{9 \sin \left( \frac{2}{\sqrt{3}} \right)}. \] \bigskip