Question

x^2+y^2=4
1)express each relation as an explicit function of x.
2) derivative

Answer 1

i.e. "solve for \(y\)"
\[x^2+y^2=4\] a circle with center \((0,0)\) and radius \(2\) so not a function, but you can solve via
\[y^2=4-x^2\]
\[y=\pm\sqrt{4-x^2}\]

Answer 2

I got that but I don't know how to do the second part

Answer 3

You will have two functions
so you will have two derivatives, one for y>=0 and the other for y<=0.

Answer 4

\[y=\sqrt{4-x^2} ; y=-\sqrt{4-x^2} \]
Find derivative of both of these by chain rule.