If $$\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0, x \neq y$$, then $$(1+x)^2 \frac{d y}{d x}=$$

If $$\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$$, then $$\frac{d y}{d x}=$$

If $$f(x)=\log (\sec x+\tan x)$$, then $$f^{\prime}\left(\frac{\pi}{4}\right)=$$