If ` y= tan^(-1) "" ((log(e //x^2))/( log (ex^2)))+ tan^(-1) "" ((3+2 log x)/(1-6 log x))` then ` (d^2 y)/(dx^2)` is

Options

A. 2
B. 1
C. 0
D. `-1`

Answer

To solve the problem, we need to differentiate the given function twice. Let's break down the solution step by step.

### Step 1: Rewrite the function
Given:
\[ y = \tan^{-1} \left( \frac{\log(e/x^2)}{\log(ex^2)} \right) + \tan^{-1} \left( \frac{3 + 2 \log x}{1 - 6 \log x} \right) \]

Using the properties of logarithms, we can rewrite the logarithmic expressions:
1. \(\log(e/x^2) = \log e - \log(x^2) = 1 - 2\log x\)
2. \(\log(ex^2) = \log e + \log(x^2) = 1 + 2\log x\)

Thus, we can rewrite \(y\) as:
\[ y = \tan^{-1} \left( \frac{1 - 2\log x}{1 + 2\log x} \right) + \tan^{-1} \left( \frac{3 + 2\log x}{1 - 6\log x} \right) \]

### Step 2: Differentiate \(y\) with respect to \(x\)
To find \(\frac{dy}{dx}\), we will use the derivative of the arctangent function:
\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

#### For the first term:
Let \(u_1 = \frac{1 - 2\log x}{1 + 2\log x}\).
Using the quotient rule:
\[ \frac{du_1}{dx} = \frac{(1 + 2\log x)(-2/x) - (1 - 2\log x)(2/x)}{(1 + 2\log x)^2} \]
Simplifying:
\[ \frac{du_1}{dx} = \frac{-2(1 + 2\log x) + 2(1 - 2\log x)}{x(1 + 2\log x)^2} = \frac{-4\log x}{x(1 + 2\log x)^2} \]

So,
\[ \frac{dy_1}{dx} = \frac{1}{1 + u_1^2} \cdot \frac{du_1}{dx} \]

#### For the second term:
Let \(u_2 = \frac{3 + 2\log x}{1 - 6\log x}\).
Using the quotient rule:
\[ \frac{du_2}{dx} = \frac{(1 - 6\log x)(2/x) - (3 + 2\log x)(-6/x)}{(1 - 6\log x)^2} \]
Simplifying:
\[ \frac{du_2}{dx} = \frac{(2 - 12\log x + 18 + 12\log x)}{x(1 - 6\log x)^2} = \frac{20}{x(1 - 6\log x)^2} \]

So,
\[ \frac{dy_2}{dx} = \frac{1}{1 + u_2^2} \cdot \frac{du_2}{dx} \]

### Step 3: Combine the derivatives
Now, we combine the derivatives:
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]

### Step 4: Differentiate again to find \(\frac{d^2y}{dx^2}\)
To find the second derivative, we differentiate \(\frac{dy}{dx}\) again using the product and chain rules.

### Final Result
The final expression for \(\frac{d^2y}{dx^2}\) will involve substituting the expressions for \(\frac{dy_1}{dx}\) and \(\frac{dy_2}{dx}\) and applying the differentiation rules accordingly.