$\frac{d}{dx}\left[ \frac{2}{\pi }\sin {x^\circ} \right] = $

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. We say that $f$ has $PROPERTY \ 1$ if $\lim_{h \rightarrow 0} \frac{f(h)-f(0)}{\sqrt{|h|}}$ exists and is finite,and $PROPERTY \ 2$ if $\lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h^2}$ exists and is finite. Then which of the following options is/are correct?
$(1) \ f(x)=x|x|$ has $PROPERTY \ 2$
$(2) \ f(x)=x^{2/3}$ has $PROPERTY \ 1$
$(3) \ f(x)=\sin x$ has $PROPERTY \ 2$
$(4) \ f(x)=|x|$ has $PROPERTY \ 1$

The derivative of $y = (1 - x)(2 - x)...(n - x)$ at $x = 1$ is equal to

Find the derivative: $\frac{d}{dx} \cosh^{-1}(\sec x) = $

If $y = \sin [\cos (\sin x)],$ then $dy/dx = $

If $y = \cos (\sin {x^2}),$ then at $x = \sqrt {\frac{\pi }{2}} , \frac{dy}{dx} = $