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

Consider the following statements:
Statement $1$: If $y = \log_{10} x + \log_{e} x$,then $\frac{dy}{dx} = \frac{\log_{10} e}{x} + \frac{1}{x}$.
Statement $2$: $\frac{d}{dx}(\log_{10} x) = \frac{\log x}{\log 10}$ and $\frac{d}{dx}(\log_{e} x) = \frac{\log x}{\log e}$.

The derivative of $\log_8(\log_5 x)$ with respect to $x$ is:

If $\frac{d}{dx} \left( A \log \left( \frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1} \right) \right) = \frac{1}{x \sqrt{1-x^3}}$,then $AB=$

$\frac{d}{d x}(\log _{|x|} e) =$ . . . . . .

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