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 $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = \cos \theta - \cos 2\theta$ and $y = \sin \theta - \sin 2\theta$.

$x=\cos \theta, y=\sin 5 \theta \Rightarrow (1-x^2) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}$ is equal to (in $y$)

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

If $x=a\left\{\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right\}$ and $y=a \sin \theta$, then $\frac{dy}{dx}$ is equal to

Find $\frac{dy}{dx}$,if $x = 2 \cos \theta - \cos 2 \theta$ and $y = 2 \sin \theta - \sin 2 \theta$.