Answer
(a)$\frac{dy}{dx}= 2\sqrt y \cos x$
(b) $\frac{dy}{dx}= 2\sin x \cos x+4\cos x$
(c)$\frac{dy}{dx}=2\sin x \cos x +4\cos x$
Work Step by Step
$\sqrt y -\sin x=2$
(a)$\frac{d}{dx}( \sqrt y -\sin x)= \frac{d}{dx}(2) $
$\frac{d}{dx} \sqrt y-\frac{d}{dx} \sin x =0 $
$ \frac{d}{dx}(y^{ \frac{1}{2}})-\cos x=0 $
$ \frac{d}{dy}(y^{ \frac{1}{2}}) \frac{dy}{dx}-\cos x=0 $
$ \frac{1}{2}y^{ \frac{1}{2}-1} \frac{dy}{dx}-\cos x=0 $
$ \frac{1}{2}y^{ -\frac{1}{2}} \frac{dy}{dx}-\cos x=0 $
$ \frac{1}{2} \frac{1}{y^{\frac{1}{2}}}\frac{dy}{dx}=\cos x $
$\frac{1}{2} \frac{1}{\sqrt y} \frac{dy}{dx}=\cos x$
$\frac{dy}{dx}= 2\sqrt y \cos x$
(b)$\sqrt y -\sin x=2$
$\sqrt y =2+\sin x$
Squaring
$(\sqrt y)^2=(2+\sin x)^2$
$y=4+\sin^2{x+4\sin x}$
$ \frac{dy}{dx}= \frac{d}{dx}( 4+\sin^2{x+4\sin x} ) $
$\frac{dy}{dx}=\frac{d}{dx}(4)+ \frac{d}{dx}
(\sin^2{x})+4\frac{d}{dx}(\sin x) $
$\frac{dy}{dx}= 0+2\sin x\frac{d}{dx}(\sin {x})+4\cos x$
$\frac{dy}{dx}= 2\sin x \cos x+4\cos x$
$\frac{dy}{dx}= 2\sin x \cos x+4\cos x$
(c)Given equation
$\sqrt y -\sin x=2$
$\sqrt y=2 +\sin x$
Putting in
$\frac{dy}{dx}= 2\sqrt y \cos x$
$\frac{dy}{dx}= 2( 2 +\sin x ) \cos x$
$\frac{dy}{dx}=4\cos x+2\sin x \cos x$
$\frac{dy}{dx}=2\sin x \cos x +4\cos x$
