If $f(x)=\log _e\left(e^{2 x}\left(\frac{3 x+5}{5-3 x}\right)^{2 / 3}\right), x \neq \frac{-5}{3}, \frac{5}{3}$, then the value of $\frac{d f}{d x}$ at $x=1$, is

If $x=\operatorname{cosec} \theta-\sin \theta, y=\operatorname{cosec}^{2022} \theta-\sin ^{2022} \theta$ and $\left(\frac{d y}{d x}\right)^2=\frac{k\left(y^2+4\right)}{g(x)}$ where $k \in R$, then $10+k-g(2022)=$

$$ \frac{d}{d x}\left[\operatorname{cosech}^{-1}(\tan 2 x)\right]= $$

Let $f: R \rightarrow R$ be defined by $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all $x$ and $y$. If $f^{\prime}(0)$ exists and equals -1 and $f(0)=1$, then $f(2)=$