If ` y=tan ^(-1) ((log (ex))/( log ((e)/( x)))) ,then (dy)/(dx) =`

Options

A. ` (1)/( x( 1+(log x) ^(2)))`
B. ` (-1)/( x( 1+(log x) ^(2)))`
C. ` (x)/( x( 1+(log x) ^(2)))`
D. ` (-x)/( x( 1+(log x) ^(2)))`

Answer

To solve the problem, we need to differentiate the function given by:

\[ y = \tan^{-1} \left( \frac{\log e x}{\log \frac{e}{x}} \right) \]

### Step 1: Simplify the expression inside the arctangent

Using the properties of logarithms, we can simplify the expression:

1. The numerator:
\[ \log e x = \log e + \log x = 1 + \log x \]
(since \(\log e = 1\))

2. The denominator:
\[ \log \frac{e}{x} = \log e - \log x = 1 - \log x \]

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

### Step 2: Use the tangent addition formula

Next, we can express this in terms of a tangent function:
\[ y = \tan^{-1} \left( \frac{\tan \frac{\pi}{4} + \tan \theta}{1 - \tan \frac{\pi}{4} \tan \theta} \right) \]
where \(\theta = \tan^{-1}(\log x)\). This follows from the identity:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]

Thus, we have:
\[ y = \tan^{-1} \left( \tan \left( \frac{\pi}{4} + \tan^{-1}(\log x) \right) \right) \]

### Step 3: Simplify further

Since \(\tan^{-1}(\tan z) = z\) for \(z\) in the principal range, we can simplify \(y\) to:
\[ y = \frac{\pi}{4} + \tan^{-1}(\log x) \]

### Step 4: Differentiate \(y\)

Now, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = 0 + \frac{d}{dx} \left( \tan^{-1}(\log x) \right) \]

Using the chain rule:
\[ \frac{d}{dx} \left( \tan^{-1}(u) \right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \(u = \log x\).

### Step 5: Calculate \(\frac{du}{dx}\)

We know:
\[ \frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x} \]

### Step 6: Substitute back into the derivative

Now substituting back, we have:
\[ \frac{dy}{dx} = \frac{1}{1 + (\log x)^2} \cdot \frac{1}{x} \]

### Final Result

Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{1}{x(1 + (\log x)^2)} \]