---

Solution

Let $a=\sqrt{x+y},\,b=\sqrt{y-x}$. $(a+b)^2=2 \Rightarrow y+\sqrt{y^{2}-x^{2}}=1$. Differentiate: $y'+\dfrac{yy'-x}{\sqrt{y^{2}-x^{2}}}=0$. But $\sqrt{y^{2}-x^{2}}=1-y$ from above ⇒ $y'=x$ ⇒ $y''=1$.

---

Solution

Differentiate both sides: $\dfrac{1}{2\sqrt{x + y}}(1 + \dfrac{dy}{dx}) + \dfrac{1}{2\sqrt{y - x}}(\dfrac{dy}{dx} - 1) = 0$. Simplify to get $\dfrac{dy}{dx} = \dfrac{\sqrt{y - x} - \sqrt{x + y}}{\sqrt{y - x} + \sqrt{x + y}}$. Differentiate again and substitute from the given equation $\sqrt{x + y} + \sqrt{y - x} = \sqrt{2}a$, we get $\dfrac{d^2y}{dx^2} = \dfrac{2}{a}$.

---

Solution

Let $u=\dfrac{x-y}{x+y}$. Since $\sec u=a$ (constant), $u' = 0$. $\displaystyle 0=\frac{d}{dx}!\left(\frac{x-y}{x+y}\right)=\frac{(1-y')(x+y)-(x-y)(1+y')}{(x+y)^2}$ $\Rightarrow 2y-2xy'=0\Rightarrow y'=\dfrac{y}{x}.$

---

Solution

$\ln(x^{y}y^{x})=y\ln x+x\ln y=\ln16$ $\Rightarrow y'(\ln x+\tfrac{x}{y})+(\tfrac{y}{x}+\ln y)=0$ At $(2,2)$: $y'=\dfrac{-,(1+\ln2)}{,\ln2+1}=-1.$