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## anonymous 4 years ago 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)

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1. anonymous

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

2. anonymous

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...

3. anonymous

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

4. anonymous

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

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