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I have the expression ${\dfrac{x^2+y^2}{x+y}}=3$ and I wanna find $dy/dx$. Here's my approach:

$x^2+y^2=3x+3y$

$\implies (x^2-3x)+(y^2-3y)=0$

$\implies (2x-3)+\dfrac{dy}{dx}(2y-3)=0$

${\implies \dfrac{dy}{dx}=\dfrac{3-2x}{2y-3}}$

Wolfram Alpha gives me a very different answer. This seems to work (i.e, the tangent I found at $x=3$ from this looks right), but I'm not sure.

Przemysław Scherwentke
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clueless
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1 Answers1

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implicit differentiating of $\frac{x^2+y^2}{x+y}=3$ gives $\frac{2x+2yy')(x+y)-(x^2+y^2)(1+y')}{(x^2+y^2)^2}=0$ plugging $x^2+y^2=3(x+y)$ in this term we get $(2x+2yy')(x+y)-3(x+y)(1+y')=0$ with $x+y\ne 0$ we get $2x+2yy'-3-y'=0$ and this is your result

Dr. Sonnhard Graubner
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