If $y = \sqrt{\tan x + y}$, then $\frac{dy}{dx} = $
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For infinite nested root problems of the form $y = \sqrt{f(x) + \sqrt{f(x) + \dots}}$, which is equivalent to $y = \sqrt{f(x) + y}$, the derivative is always given by the standard shortcut formula: $y' = \frac{f'(x)}{2y - 1}$. Here $f(x) = \tan x$, so $y' = \frac{\sec^2 x}{2y-1}$.