Answer
a) $\frac{dy}{dx} = \frac{1-y}{x+1}$
b) $\frac{dy}{dx} = \frac{1}{(x+1)^{2}}$
c) $\frac{dy}{dx} = \frac{1}{(x+1)^{2}}$
Work Step by Step
a) $xy = x-y$
Differentiating implicitly :
$x\frac{dy}{dx} + y(1) = 1-\frac{dy}{dx}$
$x\frac{dy}{dx} + \frac{dy}{dx} = 1-y$
$(x+1)\frac{dy}{dx} = 1-y$
$\frac{dy}{dx} = \frac{1-y}{x+1}$
b) $y(x+1) = x$
$y = \frac{x}{x+1}$
Differentiating :
$\frac{dy}{dx} = \frac{(x+1)(1)-x(1)}{(x+1)^{2}}$
$\frac{dy}{dx} = \frac{x+1-x}{(x+1)^{2}}$
$\frac{dy}{dx} = \frac{1}{(x+1)^{2}}$
c) $\frac{dy}{dx} = \frac{1-y}{x+1}$
As $y = \frac{x}{x+1}$
So,
$\frac{dy}{dx} = \frac{1-\frac{x}{x+1}}{x+1}$
After simplification :
$\frac{dy}{dx} = \frac{x+1-x}{(x+1)^{2}}$
$\frac{dy}{dx} = \frac{1}{(x+1)^{2}}$
