If $\sqrt{y + x} + \sqrt{y - x} = c$, then $\frac{dy}{dx} = f(x) - \sqrt{[f(x)]^2 - 1}$. Find $f(x)$.

Let $f$ be a differentiable function such that $(x - y) f(x + y) - (x + y) f(x - y) = 4xy(x^2 - y^2)$ and $f(1) = 2$. Then the area enclosed by $\frac{|f(x) - x|^{1/3}}{17} + \frac{|f(y) - y|^{1/3}}{2} \le \frac{1}{4}$ is

If $x^{y} + y^{x} = a^{b}$,then $\frac{dy}{dx}$ at $x = 1, y = 2$ is

If $x = y\sqrt{1 - y^2}$,then $\frac{dy}{dx} = $

If $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$,then $\frac{dy}{dx}$ is equal to

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