Answer
a) $\frac{dy}{dx} = 2\sqrt y$ $cosx$
b) $\frac{dy}{dx} = 4cosx + 2sinxcosx$
c) $\frac{dy}{dx} = 4cosx + 2sinxcosx$
Work Step by Step
a) $\sqrt y$ $- sinx= 2$
Taking derivative on both sides:
$\frac{d}{dx}(\sqrt y)$ $- \frac{d}{dx}(sinx) = \frac{d}{dx}(2)$
$\frac{1}{2}y^{-\frac{1}{2}}\times\frac{dy}{dx} - cosx = 0$
$\frac{1}{2}y^{-\frac{1}{2}}\times\frac{dy}{dx} = cosx$
$\frac{1}{2\sqrt y}\frac{dy}{dx} = cosx$
$\frac{dy}{dx} = 2\sqrt y$ $cosx$
b) $\sqrt y$ $- sinx= 2$
Solving for y :
$\sqrt y = 2 + sinx$
Taking square on both sides
$y = (2+sinx)^{2}$
$y = 4 + 4sinx + sin^{2}x$
Taking derivative on both sides
$\frac{dy}{dx} = 0 + 4cosx + 2sinxcosx$
So, $\frac{dy}{dx} = 4cosx + 2sinxcosx$
c) From part a)
$\frac{dy}{dx} = 2\sqrt y$ $cosx$
Putting $(y = (2+sinx)^{2})$ from part b :
$\frac{dy}{dx} = 2cosx\sqrt (2+sinx)^{2}$
$\frac{dy}{dx} = 2cosx (2+sinx)$
$\frac{dy}{dx} = 4cosx + 2sinxcosx$
