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If $x \sqrt{1+y}+y \sqrt{1+x}=0$, prove that $(1+x^2)\frac{dy}{dx}+1=0.$

The answer I got is $$\frac{dy}{dx}= -\frac{2 \sqrt{1+x} \sqrt{1+y}+y}{x+2 \sqrt{1+x}\sqrt{1+y}}$$ but I cannot simplify it further.
Please provide your assistance.

chndn
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  • http://math.stackexchange.com/questions/432514/derivative-of-x-sqrt1yy-sqrt1x-0 – lab bhattacharjee Aug 25 '13 at 14:02
  • Divide by $xy$ and you get an easier $ \dfrac{\sqrt{1+y}}{y} + \dfrac{\sqrt{1+x}}{x}=0 $ to differentiate because first part has its coefficient as derivative. – Narasimham Apr 17 '18 at 13:17
  • Another case of the older question being marked as the duplicate of the newer question. I'm sure you'll get to 5 just fine without me. – Robert Soupe Apr 17 '18 at 18:29

2 Answers2

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$$x\sqrt{1+y} = -y\sqrt{1+x}$$ squaring both sides $$ x^2(1+y) = y^2(1+x)$$ simplifying $$x^2 - y^2 = xy(y - x)$$ $$x+y = -xy$$ $$y =\frac{-x}{1+x}$$ further take derivatives.

Suraj M S
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the points $(0,0)$ and $(-1,-1)$ no longer lie on the graph. Btw on actual differentiation it yield $\frac{dy}{dx} = \frac{y^2 - 2x(1+y)}{x^2 - 2y(1+x)}$.

TZakrevskiy
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keygenx
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    In order to complete your answer, it would be better to elaborate on why you think that those point are no longer on the graph. In addition, it's better to use MathJAx to improve readabilty. Finally, it's unclear how your expression for derivative contradicts the desired result. – TZakrevskiy Jan 30 '14 at 13:03
  • sorry can't use mathjax since using mobile. (0,0) and (-1,-1) satisfied the initial equation but no longer satisfy after removing (y-x) term. Now on differentiating x + y + xy = 0 we get y^2/x^2 making substitution from original equation and -1/(1+x)^2 when substituting from x + y + xy = 0. My point is removing(y-x) alters the derivative. So is it mathematically correct? – keygenx Jan 30 '14 at 13:27
  • The function $y(x)=x$ is not a solution to the original equation, so we can safely divide by $y-x$ wherever the it's not division by zero. – TZakrevskiy Jan 30 '14 at 13:40