Questions in differentiation

Select	Question
	If $\cos x = \frac{1}{{\sqrt {1 + {t^2}} }}$and $\sin y = \frac{t}{{\sqrt {1 + {t^2}} }}$, then $\frac{{dy}}{{dx}} = $
	If$x = a(\cos \theta + \theta \sin \theta )$, $y = a(\sin \theta - \theta \cos \theta ),{\rm{ }}$then $\frac{{dy}}{{dx}} = $
	If $x = a{\cos ^4}\theta ,y = a{\sin ^4}\theta ,$ then $\frac{{dy}}{{dx}}$, at $\theta = \frac{{3\pi }}{4}$, is
	If $x = {\sin ^{ - 1}}(3t - 4{t^3})$ and $y = {\cos ^{ - 1}}\,\,\sqrt {(1 - {t^2})} $, then $\frac{{dy}}{{dx}}$ is equal to
	If $x = a\left( {t - \frac{1}{t}} \right)\,,y = a$ $\left( {t + \frac{1}{t}} \right)$then $\frac{{dy}}{{dx}} = $
	If $x = \sin t\cos 2t$ and $y = \cos t\sin 2t$, then at $t = \frac{\pi }{4},$ the value of $\frac{{dy}}{{dx}}$ is equal to
	If $\ln \,(x + y) = 2xy,$then $y'(0)$=
	If $y = {x^x}$, then $\frac{{dy}}{{dx}} = $
	The first derivative of the function $\left[ {{{\cos }^{ - 1}}\left( {\sin \sqrt {\frac{{1 + x}}{2}} } \right) + {x^x}} \right]$ with respect to x at x = 1 is
	If $y = \sqrt {\frac{{1 + x}}{{1 - x}}} ,$then $\frac{{dy}}{{dx}} = $