The value of $$\log_e 2 \cdot \dfrac{d}{dx}(\log_{\cos x} \csc x)$$ at $$x = \dfrac{\pi}{4}$$ is

We need to find the value of $$\log_e 2 \cdot \dfrac{d}{dx}(\log_{\cos x} \csc x)$$ at $$x = \dfrac{\pi}{4}$$.

$$\log_{\cos x} \csc x = \frac{\ln(\csc x)}{\ln(\cos x)}$$

Let $$u = \ln(\csc x)$$ and $$v = \ln(\cos x)$$.

$$u' = \frac{-\csc x \cot x}{\csc x} = -\cot x$$

$$v' = \frac{-\sin x}{\cos x} = -\tan x$$

$$f'(x) = \frac{u'v - uv'}{v^2} = \frac{-\cot x \cdot \ln(\cos x) - \ln(\csc x) \cdot (-\tan x)}{[\ln(\cos x)]^2}$$

$$= \frac{-\cot x \cdot \ln(\cos x) + \tan x \cdot \ln(\csc x)}{[\ln(\cos x)]^2}$$

At $$x = \frac{\pi}{4}$$:

$$\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \csc\frac{\pi}{4} = \sqrt{2}$$

$$\cot\frac{\pi}{4} = 1, \quad \tan\frac{\pi}{4} = 1$$

$$\ln(\cos\frac{\pi}{4}) = \ln\frac{1}{\sqrt{2}} = -\frac{1}{2}\ln 2$$

$$\ln(\csc\frac{\pi}{4}) = \ln\sqrt{2} = \frac{1}{2}\ln 2$$

Numerator: $$-1 \cdot (-\frac{1}{2}\ln 2) + 1 \cdot \frac{1}{2}\ln 2 = \frac{1}{2}\ln 2 + \frac{1}{2}\ln 2 = \ln 2$$

Denominator: $$\left(-\frac{1}{2}\ln 2\right)^2 = \frac{1}{4}(\ln 2)^2$$

$$f'\left(\frac{\pi}{4}\right) = \frac{\ln 2}{\frac{1}{4}(\ln 2)^2} = \frac{4}{\ln 2} = \frac{4}{\log_e 2}$$

$$\log_e 2 \cdot f'\left(\frac{\pi}{4}\right) = \log_e 2 \cdot \frac{4}{\log_e 2} = 4$$

Therefore, the correct answer is Option D: $$4$$.