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Find $y'=\frac{dy}{dx}$ when: $$2x^2-xy+y^2=1$$

How do I solve this?
Do I start with this?:

$$y=\frac{2x^2+y^2-1}{x}$$

$$y'=\left(\frac{2x^2+y^2-1}{x}\right)'$$

Aditya Hase
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peppa
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1 Answers1

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You can follow in this trace:

$$(2x^2-xy+y^2)'=1' \implies 4x-(y+xy')+2yy'=0 \implies y'=\frac{y-4x}{2y-x}.$$

Paul
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  • Why is: $$(xy)'=y+xy'$$ – peppa Nov 19 '14 at 03:43
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    You should view $y$ as a function on $x$. @Peppa, Could you upvote me so that I will get 10k reputation:) – Paul Nov 19 '14 at 03:45
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    @peppa First, remember that $(fg)'=f'g+fg'$ (the product rule). Also remember that $x'=\frac{\operatorname d!x}{\operatorname d!x}=1$. Now, use the product rule, where $f=x$ and $g=y$. – Akiva Weinberger Nov 19 '14 at 04:18