Question

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

• A. $\frac{2}{1+x^2}$
• B. $\frac{1}{1+x^2}$
• C. $\frac{3}{1+x^2}$
• D. $0$

Hint:

Whenever an inverse trigonometric expression looks long and intimidating, look for hidden algebraic cancellations using standard angle addition identities like $\tan^{-1}\left(\frac{A \pm B}{1 \mp AB}\right)$. These often reduce complex variable expressions down to a simple constant angle, meaning the derivative is instantly zero!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem asks for the second derivative $\frac{d^2y}{dx^2}$ of a given sum of two complex inverse trigonometric functions involving logarithmic expressions.

Step 2: Key Formula or Approach:
We simplify the arguments of the inverse tangents before differentiating by applying standard logarithmic properties: $$\log\left(\frac{A}{B}\right) = \log A - \log B \quad \text{and} \quad \log(AB) = \log A + \log B$$ along with the inverse tangent addition identity: $$\tan^{-1}\left(\frac{A - B}{1 + AB}\right) = \tan^{-1}A - \tan^{-1}B$$ $$\tan^{-1}\left(\frac{A + B}{1 - AB}\right) = \tan^{-1}A + \tan^{-1}B$$

Step 3: Detailed Explanation:
Let's first expand the terms inside the first inverse tangent function using $\log e = 1$: $$\log\left(\frac{e}{x^2}\right) = \log e - \log x^2 = 1 - \log x^2$$ $$\log(ex^2) = \log e + \log x^2 = 1 + \log x^2$$ Substitute these expansions back into the first term of our function: $$y_1 = \tan^{-1}\left[\frac{1 - \log x^2}{1 + 1 \cdot \log x^2}\right] = \tan^{-1}(1) - \tan^{-1}(\log x^2)$$ Now, look at the second inverse tangent term. Notice that the denominator can be rewritten to match our identity format where $A = 3$ and $B = 2\log x$ (since $AB = 3 \cdot 2\log x = 6\log x$): $$y_2 = \tan^{-1}\left[\frac{3 + 2\log x}{1 - 3 \cdot (2\log x)}\right] = \tan^{-1}(3) + \tan^{-1}(2\log x)$$ Using the power rule of logarithms, we know that $\log x^2 = 2\log x$. Substituting this back into our expression for $y_1$ gives: $$y_1 = \tan^{-1}(1) - \tan^{-1}(2\log x)$$ Now combine the two parts together to get the full simplified expression for $y$: $$y = y_1 + y_2 = \left[\tan^{-1}(1) - \tan^{-1}(2\log x)\right] + \left[\tan^{-1}(3) + \tan^{-1}(2\log x)\right]$$ The variable terms $-\tan^{-1}(2\log x)$ and $+\tan^{-1}(2\log x)$ cancel each other out completely: $$y = \tan^{-1}(1) + \tan^{-1}(3)$$ Since $y$ is equal to a sum of constant angles, it behaves as a constant value with respect to $x$. Differentiating a constant with respect to $x$ yields zero for both the first and second derivatives: $$\frac{dy}{dx} = 0 \implies \frac{d^2y}{dx^2} = 0$$

Step 4: Final Answer:
The second derivative evaluates cleanly to $0$, which matches option (D).