If $f^{\prime}(x)=\sqrt{2 x^2-1}$ and $y=f(x^3)$,then find the value of $\frac{dy}{dx}$ at $x=1$.

If $\tan u=\sqrt{\frac{1-x}{1+x}}$ and $\cos v=4 x^{3}-3 x$,then $\frac{d u}{d v}=$

If $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$,then $\frac{dy}{dx} = $

If $x=a(t-\sin t)$ and $y=a(1+\cos t)$, then $\frac{d^2 y}{d x^2}=$

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{d y}{d x}$:
$x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$

If $x = a (t - 1/t)$ and $y = a (t + 1/t)$,where $t$ is the parameter,then $dy/dx = ?$