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If $\mathrm{x} \sqrt{1+\mathrm{y}}+\mathrm{y} \sqrt{1+\mathrm{x}}=0$, what is $\frac{\mathrm{dy}}{\mathrm{dx}}$ equal to?
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Verified Answer
The correct answer is:
$-\frac{1}{(1+x)^{2}}$
Given equation
$\mathrm{x} \sqrt{1+\mathrm{y}}+\mathrm{y} \sqrt{1+\mathrm{x}}=0$
Can be written as:
$\mathrm{x} \sqrt{1+\mathrm{y}}=-\mathrm{y} \sqrt{1+\mathrm{x}}$
Squaring both sides, we get $\begin{aligned} & x^{2}(1+y)=y^{2}(1+x) \\ \Rightarrow & x^{2}+x^{2} y=y^{2}+y^{2} x \quad \Rightarrow \quad x^{2}-y^{2}=y^{2} x-x^{2} y \\ \Rightarrow &(x-y)(x+y)=-x y(x-y) \\ \Rightarrow & x+y=-x y \quad \Rightarrow y(1+x)=-x \end{aligned}$
$\mathrm{y}=\frac{-\mathrm{x}}{1+\mathrm{x}}$ which is in explicit form.
Differentiating w.r.t.x, we get
$\frac{d y}{d x}=\frac{(1+x)(-1)+x(1)}{(1+x)^{2}}=\frac{-1}{(1+x)^{2}}$
$\mathrm{x} \sqrt{1+\mathrm{y}}+\mathrm{y} \sqrt{1+\mathrm{x}}=0$
Can be written as:
$\mathrm{x} \sqrt{1+\mathrm{y}}=-\mathrm{y} \sqrt{1+\mathrm{x}}$
Squaring both sides, we get $\begin{aligned} & x^{2}(1+y)=y^{2}(1+x) \\ \Rightarrow & x^{2}+x^{2} y=y^{2}+y^{2} x \quad \Rightarrow \quad x^{2}-y^{2}=y^{2} x-x^{2} y \\ \Rightarrow &(x-y)(x+y)=-x y(x-y) \\ \Rightarrow & x+y=-x y \quad \Rightarrow y(1+x)=-x \end{aligned}$
$\mathrm{y}=\frac{-\mathrm{x}}{1+\mathrm{x}}$ which is in explicit form.
Differentiating w.r.t.x, we get
$\frac{d y}{d x}=\frac{(1+x)(-1)+x(1)}{(1+x)^{2}}=\frac{-1}{(1+x)^{2}}$
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