Question

Did I do anything wrong with this?
x^4-x^2+y^2=0, find y'(x)

x^4-x^2+f(x)^2=0
4x^3-2x+2f(x)*f'(x)=0
(2x-4x^3)/2f(x)=f'(x)
y'(x)=(x-2x^3)/f(x)

Answer 1

that's right but you can put f(x) into.

Answer 2

hmmm... weird. When I try to graph x^4-x^2+y^2=0 and y=((0.5-2(0.5^3))/0.433013)+0.433013, it doesn't seem to work...

Answer 3

\[4x^3-2x+y*y'=0\]
\[y'=(4x^3-2x)/y\]
\[y=\sqrt{-x^4+x^2}\]
\[y'=\frac{4x^3-2x}{\sqrt{-x^4-2x}}\]

Answer 4

\[y'=\frac{4x^3-2x}{\sqrt{-x^4+x^2}}\]