If `xlog_e(log_ex)-x^2+y^2=4(ygt0)",then" dy//dx" at " x=e` is equal to

Options

A. `e/(4+e^2)`
B. `((1+2e))/(2sqrt(4+e^2))`
C. `((2e-1))/(2sqrt(4+e^2))`
D. `((1+2e))/(sqrt(4+e^2))`

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) for the equation given:

\[ x \log_e(\log_e x) - x^2 + y^2 = 4 \quad (y > 0) \]

at the point \(x = e\).

### Step 1: Differentiate the equation implicitly
We start by differentiating both sides of the equation with respect to \(x\).

\[ \frac{d}{dx}\left(x \log_e(\log_e x)\right) - \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0 \]

Using the product rule on the first term:

\[ \frac{d}{dx}(x \log_e(\log_e x)) = \log_e(\log_e x) + x \cdot \frac{1}{\log_e x} \cdot \frac{1}{x} - 2x + 2y \frac{dy}{dx} \]

So we have:

\[ \log_e(\log_e x) + \frac{1}{\log_e x} - 2x + 2y \frac{dy}{dx} = 0 \]

### Step 2: Rearranging the equation
Rearranging gives us:

\[ 2y \frac{dy}{dx} = 2x - \log_e(\log_e x) - \frac{1}{\log_e x} \]

Thus,

\[ \frac{dy}{dx} = \frac{2x - \log_e(\log_e x) - \frac{1}{\log_e x}}{2y} \]

### Step 3: Substitute \(x = e\)
Now we substitute \(x = e\) into our equation to find \(y\):

\[ e \log_e(\log_e e) - e^2 + y^2 = 4 \]

Since \(\log_e e = 1\):

\[ e \cdot 1 - e^2 + y^2 = 4 \]

This simplifies to:

\[ e - e^2 + y^2 = 4 \]

Rearranging gives:

\[ y^2 = 4 + e^2 - e \]

### Step 4: Substitute \(x = e\) into \(\frac{dy}{dx}\)
Now we substitute \(x = e\) and \(y = \sqrt{4 + e^2 - e}\) into our expression for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{2e - \log_e(\log_e e) - \frac{1}{\log_e e}}{2y} \]

Substituting \(\log_e(\log_e e) = 0\) and \(\frac{1}{\log_e e} = 1\):

\[ \frac{dy}{dx} = \frac{2e - 0 - 1}{2y} = \frac{2e - 1}{2y} \]

### Step 5: Final substitution for \(y\)
Substituting \(y = \sqrt{4 + e^2 - e}\):

\[ \frac{dy}{dx} = \frac{2e - 1}{2\sqrt{4 + e^2 - e}} \]

### Final Result
Thus, the value of \(\frac{dy}{dx}\) at \(x = e\) is:

\[ \frac{dy}{dx} = \frac{2e - 1}{2\sqrt{4 + e^2 - e}} \]