If $y = \log_{2026}(\log_{2025} x)$, then $\frac{dy}{dx} = \dots \dots \dots$

If $y = \log \log x$,then $e^y \frac{dy}{dx} = $

If $y = \log \left(\frac{1-x^{2}}{1+x^{2}}\right)$,then $\frac{dy}{dx}$ is equal to

If $y = \log^n x$, where $\log^n$ denotes the $n$-th iterated logarithm $\log_e(\log_e(\dots \log_e x \dots))$ ($n$ times), then $x \log x \log^2 x \log^3 x \dots \log^{n-1} x \log^n x \frac{dy}{dx}$ is equal to

$f(x) = x^{x}$ has a stationary point at

If $y = e^{(1 + \log_e x)}$,then the value of $\frac{dy}{dx} = $