Find $\frac{dy}{dx}$ if $x(t) = \sqrt{\cos{t}}$ and $y(t) =\sqrt{\sin{t}}$ with $t \in (0, \pi/2)$ 
Find $\frac{dy}{dx}$ if $x(t) = \sqrt{\cos{t}}$ and $y(t) =\sqrt{\sin{t}}$ with $t \in (0, \pi/2)$

I found that $\frac{dy}{dx} = -\frac{1}{4}\sqrt{\sin{t}}\sqrt{\cos{t}} = -\frac{1}{4}x(t)y(t). $ However, I was asked to express everything in terms of $x$ and $y$. I'm confused because I'm not sure if that's the same thing as $x(t)$ and $y(t)$. 
 A: if you want to express $\frac{dy}{dx}$ in terms of $x, y$, then why don't you use the fact that $$x^4 + y^4 = 1 .$$differentiating the above gives you $$x^3+y^3\frac{dy}{dx} = 0. $$
A: Note that we have 
$$
x^4+y^4
= \left(\sqrt{\cos t}\right)^4+\left(\sqrt{\sin t}\right)^4 
= \cos^2t+\sin^2t 
= 1\tag{1}
$$
Differentiating (1) with respect to $x$ gives
$$
4\,x^3+4\,y^3\frac{dy}{dx}=0\tag{2}
$$
Solving (2) for $dy/dx$ gives
$$
\frac{dy}{dx}=-\frac{x^3}{y^3}
$$
Note that this formula is valid on the entire specified domain.
A: You should get $$\frac{dy}{dx}=-\frac{x^3}{y^3}$$
Note that, as an alternative method, $x^4+y^4=1$, so the result can be found also by implicit differentiation
A: You can even do it directly since $$x = \sqrt{\cos{t}}\implies \frac{dx}{dt}=-\frac{\sin (t)}{2 \sqrt{\cos (t)}}$$ $$y =\sqrt{\sin{t}}\implies \frac{dy}{dt}=\frac{\cos (t)}{2 \sqrt{\sin (t)}}$$ $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=-\frac{\cos ^{\frac{3}{2}}(t)}{\sin ^{\frac{3}{2}}(t)}=-\frac{x^3}{y^3}$$
