If $$x \log_e(\log_e x) - x^2 + y^2 = 4$$ $$(y > 0)$$, then $$\frac{dy}{dx}$$ at $$x = e$$ is equal to:

We have the implicit relation

$$x\,\log_e\!\bigl(\log_e x\bigr)\;-\;x^{2}\;+\;y^{2}\;=\;4.$$

Here $$y$$ is a function of $$x$$ and we have to find $$\dfrac{dy}{dx}$$ at $$x=e$$, given that $$y>0$$.

First we differentiate both sides with respect to $$x$$. Whenever we differentiate a term containing $$y$$ we must multiply by $$\dfrac{dy}{dx}$$ (chain rule).

The derivative of the constant $$4$$ is $$0$$, so

$$\frac{d}{dx}\Bigl[x\log_e\!\bigl(\log_e x\bigr)\Bigr]\;-\;\frac{d}{dx}(x^{2})\;+\;\frac{d}{dx}(y^{2})\;=\;0.$$

Now we handle each derivative separately.

1. Derivative of $$x\log_e(\log_e x)$$. We use the product rule $$\dfrac{d}{dx}(uv)=u'\,v+u\,v'$$ with $$u=x,\;v=\log_e(\log_e x).$$ We have $$u'=1$$ and $$v'=\dfrac{1}{\log_e x}\cdot\dfrac{1}{x}=\dfrac{1}{x\log_e x}.$$ So

$$\frac{d}{dx}\Bigl[x\log_e(\log_e x)\Bigr]=1\cdot\log_e(\log_e x)+x\cdot\dfrac{1}{x\log_e x}=\log_e(\log_e x)+\frac{1}{\log_e x}.$$

2. Derivative of $$-x^{2}$$. Using the power rule $$\dfrac{d}{dx}(x^{n})=nx^{n-1}$$ we get

$$\frac{d}{dx}(-x^{2})=-2x.$$

3. Derivative of $$y^{2}$$. Using the chain rule $$\dfrac{d}{dx}(y^{2})=2y\dfrac{dy}{dx}.$$

Putting the three results together:

$$\Bigl[\log_e(\log_e x)+\frac{1}{\log_e x}\Bigr]\;-\;2x\;+\;2y\,\frac{dy}{dx}=0.$$

We now isolate $$\dfrac{dy}{dx}$$.

$$2y\,\frac{dy}{dx}=2x-\log_e(\log_e x)-\frac{1}{\log_e x},$$

$$\frac{dy}{dx}=\frac{2x-\log_e(\log_e x)-\dfrac{1}{\log_e x}}{2y}.$$

Next we substitute $$x=e$$.

Since $$\log_e e=1,$$ we obtain $$\log_e x=1$$ and therefore $$\log_e(\log_e x)=\log_e 1=0.$$ Substituting these values gives

$$\frac{dy}{dx}\Big|_{x=e}=\frac{2e-0-\dfrac{1}{1}}{2y}=\frac{2e-1}{2y}.$$

To complete the evaluation, we must find $$y$$ when $$x=e$$. We go back to the original equation and put $$x=e$$.

$$e\,\log_e(\log_e e)\;-\;e^{2}\;+\;y^{2}=4.$$

Again $$\log_e e=1$$ and $$\log_e 1=0$$, so the first term is $$0$$. Hence

$$-e^{2}+y^{2}=4,$$

$$y^{2}=4+e^{2}.$$

Because $$y>0$$, we take the positive square root:

$$y=\sqrt{\,4+e^{2}\,}.$$

Finally, substituting this value of $$y$$ into our derivative:

$$\frac{dy}{dx}\Big|_{x=e}=\frac{2e-1}{2\sqrt{\,4+e^{2}\,}}.$$

Therefore, $$\frac{dy}{dx}\Big|_{x=e}=\frac{(2e-1)}{2\sqrt{4+e^{2}}}.$$

Hence, the correct answer is Option B.