If $(\mathrm{a}+\mathrm{b} x) \mathrm{e}^{\frac{y}{x}}=x$, then $x^3 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}$ is equal to

If $y=\frac{\mathrm{K}^{\cos ^{-1} x}}{1+\mathrm{K}^{\cos ^{-1} x}}$ and $\mathrm{t}=\mathrm{K}^{\cos ^{-1} x}$, then $\frac{\mathrm{d} y}{\mathrm{dt}}=$

If $\mathrm{a}\left(4+x^2\right)=x$ and $y-x^3=\mathrm{a}^2$ then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1$ is $\qquad$

If $x^y+y^x=\mathrm{a}^{\mathrm{b}}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1, y=2$ is