Question

If $$ y = \tan^{-1} \left[ \frac{\log_e\left(\frac{e}{x}\right)}{\log_e(e x^2)} \right] + \tan^{-1} \left[ \frac{3 + 2\log_e x}{1 - 6\cdot\log_e x} \right], $$ then \( \frac{d^2y}{dx^2} = ? \) 

• A. 2
• B. 1
• C. 0
• D. -1

Hint:

For composite tan^−1 expressions, simplify the arguments as much as possible before differentiating. Pay attention to logarithmic terms.

Answer 1

The Correct Option is C

1. The given function is:

\[ y = \tan^{-1} \left[ \frac{\log_e \left(\frac{e}{x^2}\right)}{\log_e(e x^2)} \right] + \tan^{-1} \left[ \frac{3 + 2 \log_e x}{1 - 6 \cdot \log_e x} \right]. \]

2. Simplify the first term:

\[ \tan^{-1} \left[ \frac{\log_e \left(\frac{e}{x^2}\right)}{\log_e(e x^2)} \right] = \tan^{-1} \left[ \frac{\log_e e - 2 \log_e x}{\log_e e + 2 \log_e x} \right]. \]

Substitute \(\log_e e = 1\):

\[ = \tan^{-1} \left[ \frac{1 - 2 \log_e x}{1 + 2 \log_e x} \right]. \]

3. Simplify the second term:

\[ \tan^{-1} \left[ \frac{3 + 2 \log_e x}{1 - 6 \cdot \log_e x} \right]. \]

4. Differentiate \(y\) with respect to \(x\):

For \(\tan^{-1}(u)\), the derivative is:

\[ \frac{d}{dx} \tan^{-1}(u) = \frac{u'}{1 + u^2}. \]

5. Compute \(\frac{dy}{dx}\) for each term:

- For the first term:

\[ u = \frac{1 - 2 \log_e x}{1 + 2 \log_e x}, \quad u' = \frac{-2/x}{(1 + 2 \log_e x)^2}. \]

Thus:

\[ \frac{d}{dx} \tan^{-1} \left[ \frac{1 - 2 \log_e x}{1 + 2 \log_e x} \right] = \frac{-2/x}{(1 + 2 \log_e x)^2} \cdot \frac{1}{1 + \left( \frac{1 - 2 \log_e x}{1 + 2 \log_e x} \right)^2}. \]

- For the second term, differentiate similarly using:

\[ u = \frac{3 + 2 \log_e x}{1 - 6 \cdot \log_e x}, \quad u' = \frac{2/x (1 - 6 \cdot \log_e x) - (3 + 2 \log_e x)(-6/x)}{(1 - 6 \cdot \log_e x)^2}. \]

6. After simplifications, it is observed that the higher-order terms cancel out, and:

\[ \frac{d^2 y}{dx^2} = 0. \]

Thus, the correct answer is (C).